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How many institutions are there?
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CREATE TABLE inst (Id VARCHAR)
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SELECT COUNT(*) FROM inst
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### Context: CREATE TABLE inst (Id VARCHAR) ### Question: How many institutions are there? ### Answer: SELECT COUNT(*) FROM inst
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How many papers are published in total?
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CREATE TABLE papers (Id VARCHAR)
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SELECT COUNT(*) FROM papers
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### Context: CREATE TABLE papers (Id VARCHAR) ### Question: How many papers are published in total? ### Answer: SELECT COUNT(*) FROM papers
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What are the titles of papers published by "Jeremy Gibbons"?
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR)
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SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Jeremy" AND t1.lname = "Gibbons"
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### Context: CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR) ### Question: What are the titles of papers published by "Jeremy Gibbons"? ### Answer: SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Jeremy" AND t1.lname = "Gibbons"
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Find all the papers published by "Aaron Turon".
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR)
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SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Aaron" AND t1.lname = "Turon"
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### Context: CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR) ### Question: Find all the papers published by "Aaron Turon". ### Answer: SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Aaron" AND t1.lname = "Turon"
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How many papers have "Atsushi Ohori" published?
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR); CREATE TABLE papers (paperid VARCHAR)
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SELECT COUNT(*) FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Atsushi" AND t1.lname = "Ohori"
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### Context: CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR); CREATE TABLE papers (paperid VARCHAR) ### Question: How many papers have "Atsushi Ohori" published? ### Answer: SELECT COUNT(*) FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Atsushi" AND t1.lname = "Ohori"
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What is the name of the institution that "Matthias Blume" belongs to?
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CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR); CREATE TABLE inst (name VARCHAR, instid VARCHAR)
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SELECT DISTINCT t3.name FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t1.fname = "Matthias" AND t1.lname = "Blume"
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### Context: CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR); CREATE TABLE inst (name VARCHAR, instid VARCHAR) ### Question: What is the name of the institution that "Matthias Blume" belongs to? ### Answer: SELECT DISTINCT t3.name FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t1.fname = "Matthias" AND t1.lname = "Blume"
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Which institution does "Katsuhiro Ueno" belong to?
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CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR); CREATE TABLE inst (name VARCHAR, instid VARCHAR)
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SELECT DISTINCT t3.name FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t1.fname = "Katsuhiro" AND t1.lname = "Ueno"
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### Context: CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR); CREATE TABLE inst (name VARCHAR, instid VARCHAR) ### Question: Which institution does "Katsuhiro Ueno" belong to? ### Answer: SELECT DISTINCT t3.name FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t1.fname = "Katsuhiro" AND t1.lname = "Ueno"
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Who belong to the institution "University of Oxford"? Show the first names and last names.
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CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (fname VARCHAR, lname VARCHAR, authid VARCHAR); CREATE TABLE inst (instid VARCHAR, name VARCHAR)
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SELECT DISTINCT t1.fname, t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "University of Oxford"
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### Context: CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (fname VARCHAR, lname VARCHAR, authid VARCHAR); CREATE TABLE inst (instid VARCHAR, name VARCHAR) ### Question: Who belong to the institution "University of Oxford"? Show the first names and last names. ### Answer: SELECT DISTINCT t1.fname, t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "University of Oxford"
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Which authors belong to the institution "Google"? Show the first names and last names.
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CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (fname VARCHAR, lname VARCHAR, authid VARCHAR); CREATE TABLE inst (instid VARCHAR, name VARCHAR)
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SELECT DISTINCT t1.fname, t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Google"
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### Context: CREATE TABLE authorship (authid VARCHAR, instid VARCHAR); CREATE TABLE authors (fname VARCHAR, lname VARCHAR, authid VARCHAR); CREATE TABLE inst (instid VARCHAR, name VARCHAR) ### Question: Which authors belong to the institution "Google"? Show the first names and last names. ### Answer: SELECT DISTINCT t1.fname, t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Google"
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What are the last names of the author of the paper titled "Binders Unbound"?
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (lname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR)
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SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title = "Binders Unbound"
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### Context: CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (lname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR) ### Question: What are the last names of the author of the paper titled "Binders Unbound"? ### Answer: SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title = "Binders Unbound"
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Find the first and last name of the author(s) who wrote the paper "Nameless, Painless".
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR); CREATE TABLE authors (fname VARCHAR, lname VARCHAR, authid VARCHAR)
|
SELECT t1.fname, t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title = "Nameless , Painless"
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### Context: CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR); CREATE TABLE authors (fname VARCHAR, lname VARCHAR, authid VARCHAR) ### Question: Find the first and last name of the author(s) who wrote the paper "Nameless, Painless". ### Answer: SELECT t1.fname, t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title = "Nameless , Painless"
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What are the papers published under the institution "Indiana University"?
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CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR)
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SELECT DISTINCT t1.title FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Indiana University"
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### Context: CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR) ### Question: What are the papers published under the institution "Indiana University"? ### Answer: SELECT DISTINCT t1.title FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Indiana University"
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Find all the papers published by the institution "Google".
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CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR)
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SELECT DISTINCT t1.title FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Google"
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### Context: CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR) ### Question: Find all the papers published by the institution "Google". ### Answer: SELECT DISTINCT t1.title FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Google"
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How many papers are published by the institution "Tokohu University"?
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CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR)
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SELECT COUNT(DISTINCT t1.title) FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Tokohu University"
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### Context: CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR) ### Question: How many papers are published by the institution "Tokohu University"? ### Answer: SELECT COUNT(DISTINCT t1.title) FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "Tokohu University"
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Find the number of papers published by the institution "University of Pennsylvania".
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CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR)
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SELECT COUNT(DISTINCT t1.title) FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "University of Pennsylvania"
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### Context: CREATE TABLE inst (instid VARCHAR, name VARCHAR); CREATE TABLE authorship (paperid VARCHAR, instid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR) ### Question: Find the number of papers published by the institution "University of Pennsylvania". ### Answer: SELECT COUNT(DISTINCT t1.title) FROM papers AS t1 JOIN authorship AS t2 ON t1.paperid = t2.paperid JOIN inst AS t3 ON t2.instid = t3.instid WHERE t3.name = "University of Pennsylvania"
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Find the papers which have "Olin Shivers" as an author.
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR)
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SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Olin" AND t1.lname = "Shivers"
|
### Context: CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR) ### Question: Find the papers which have "Olin Shivers" as an author. ### Answer: SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Olin" AND t1.lname = "Shivers"
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Which papers have "Stephanie Weirich" as an author?
|
CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR)
|
SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Stephanie" AND t1.lname = "Weirich"
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### Context: CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE authors (authid VARCHAR, fname VARCHAR, lname VARCHAR) ### Question: Which papers have "Stephanie Weirich" as an author? ### Answer: SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t1.fname = "Stephanie" AND t1.lname = "Weirich"
|
Which paper is published in an institution in "USA" and have "Turon" as its second author?
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CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR, instid VARCHAR, authorder VARCHAR); CREATE TABLE authors (authid VARCHAR, lname VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE inst (instid VARCHAR, country VARCHAR)
|
SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid JOIN inst AS t4 ON t2.instid = t4.instid WHERE t4.country = "USA" AND t2.authorder = 2 AND t1.lname = "Turon"
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### Context: CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR, instid VARCHAR, authorder VARCHAR); CREATE TABLE authors (authid VARCHAR, lname VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE inst (instid VARCHAR, country VARCHAR) ### Question: Which paper is published in an institution in "USA" and have "Turon" as its second author? ### Answer: SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid JOIN inst AS t4 ON t2.instid = t4.instid WHERE t4.country = "USA" AND t2.authorder = 2 AND t1.lname = "Turon"
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Find the titles of papers whose first author is affiliated with an institution in the country "Japan" and has last name "Ohori"?
|
CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR, instid VARCHAR, authorder VARCHAR); CREATE TABLE authors (authid VARCHAR, lname VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE inst (instid VARCHAR, country VARCHAR)
|
SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid JOIN inst AS t4 ON t2.instid = t4.instid WHERE t4.country = "Japan" AND t2.authorder = 1 AND t1.lname = "Ohori"
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### Context: CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR, instid VARCHAR, authorder VARCHAR); CREATE TABLE authors (authid VARCHAR, lname VARCHAR); CREATE TABLE papers (title VARCHAR, paperid VARCHAR); CREATE TABLE inst (instid VARCHAR, country VARCHAR) ### Question: Find the titles of papers whose first author is affiliated with an institution in the country "Japan" and has last name "Ohori"? ### Answer: SELECT t3.title FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid JOIN inst AS t4 ON t2.instid = t4.instid WHERE t4.country = "Japan" AND t2.authorder = 1 AND t1.lname = "Ohori"
|
What is the last name of the author that has published the most papers?
|
CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (lname VARCHAR, fname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR)
|
SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.fname, t1.lname ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (lname VARCHAR, fname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR) ### Question: What is the last name of the author that has published the most papers? ### Answer: SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.fname, t1.lname ORDER BY COUNT(*) DESC LIMIT 1
|
Retrieve the country that has published the most papers.
|
CREATE TABLE inst (country VARCHAR, instid VARCHAR); CREATE TABLE authorship (instid VARCHAR, paperid VARCHAR); CREATE TABLE papers (paperid VARCHAR)
|
SELECT t1.country FROM inst AS t1 JOIN authorship AS t2 ON t1.instid = t2.instid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.country ORDER BY COUNT(*) DESC LIMIT 1
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### Context: CREATE TABLE inst (country VARCHAR, instid VARCHAR); CREATE TABLE authorship (instid VARCHAR, paperid VARCHAR); CREATE TABLE papers (paperid VARCHAR) ### Question: Retrieve the country that has published the most papers. ### Answer: SELECT t1.country FROM inst AS t1 JOIN authorship AS t2 ON t1.instid = t2.instid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.country ORDER BY COUNT(*) DESC LIMIT 1
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Find the name of the organization that has published the largest number of papers.
|
CREATE TABLE inst (name VARCHAR, instid VARCHAR); CREATE TABLE authorship (instid VARCHAR, paperid VARCHAR); CREATE TABLE papers (paperid VARCHAR)
|
SELECT t1.name FROM inst AS t1 JOIN authorship AS t2 ON t1.instid = t2.instid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.name ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE inst (name VARCHAR, instid VARCHAR); CREATE TABLE authorship (instid VARCHAR, paperid VARCHAR); CREATE TABLE papers (paperid VARCHAR) ### Question: Find the name of the organization that has published the largest number of papers. ### Answer: SELECT t1.name FROM inst AS t1 JOIN authorship AS t2 ON t1.instid = t2.instid JOIN papers AS t3 ON t2.paperid = t3.paperid GROUP BY t1.name ORDER BY COUNT(*) DESC LIMIT 1
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Find the titles of the papers that contain the word "ML".
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CREATE TABLE papers (title VARCHAR)
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SELECT title FROM papers WHERE title LIKE "%ML%"
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### Context: CREATE TABLE papers (title VARCHAR) ### Question: Find the titles of the papers that contain the word "ML". ### Answer: SELECT title FROM papers WHERE title LIKE "%ML%"
|
Which paper's title contains the word "Database"?
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CREATE TABLE papers (title VARCHAR)
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SELECT title FROM papers WHERE title LIKE "%Database%"
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### Context: CREATE TABLE papers (title VARCHAR) ### Question: Which paper's title contains the word "Database"? ### Answer: SELECT title FROM papers WHERE title LIKE "%Database%"
|
Find the first names of all the authors who have written a paper with title containing the word "Functional".
|
CREATE TABLE authors (fname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR); CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR)
|
SELECT t1.fname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title LIKE "%Functional%"
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### Context: CREATE TABLE authors (fname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR); CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR) ### Question: Find the first names of all the authors who have written a paper with title containing the word "Functional". ### Answer: SELECT t1.fname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title LIKE "%Functional%"
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Find the last names of all the authors that have written a paper with title containing the word "Monadic".
|
CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (lname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR)
|
SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title LIKE "%Monadic%"
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### Context: CREATE TABLE authorship (authid VARCHAR, paperid VARCHAR); CREATE TABLE authors (lname VARCHAR, authid VARCHAR); CREATE TABLE papers (paperid VARCHAR, title VARCHAR) ### Question: Find the last names of all the authors that have written a paper with title containing the word "Monadic". ### Answer: SELECT t1.lname FROM authors AS t1 JOIN authorship AS t2 ON t1.authid = t2.authid JOIN papers AS t3 ON t2.paperid = t3.paperid WHERE t3.title LIKE "%Monadic%"
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Retrieve the title of the paper that has the largest number of authors.
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CREATE TABLE authorship (authorder INTEGER); CREATE TABLE authorship (paperid VARCHAR, authorder INTEGER); CREATE TABLE papers (title VARCHAR, paperid VARCHAR)
|
SELECT t2.title FROM authorship AS t1 JOIN papers AS t2 ON t1.paperid = t2.paperid WHERE t1.authorder = (SELECT MAX(authorder) FROM authorship)
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### Context: CREATE TABLE authorship (authorder INTEGER); CREATE TABLE authorship (paperid VARCHAR, authorder INTEGER); CREATE TABLE papers (title VARCHAR, paperid VARCHAR) ### Question: Retrieve the title of the paper that has the largest number of authors. ### Answer: SELECT t2.title FROM authorship AS t1 JOIN papers AS t2 ON t1.paperid = t2.paperid WHERE t1.authorder = (SELECT MAX(authorder) FROM authorship)
|
What is the first name of the author with last name "Ueno"?
|
CREATE TABLE authors (fname VARCHAR, lname VARCHAR)
|
SELECT fname FROM authors WHERE lname = "Ueno"
|
### Context: CREATE TABLE authors (fname VARCHAR, lname VARCHAR) ### Question: What is the first name of the author with last name "Ueno"? ### Answer: SELECT fname FROM authors WHERE lname = "Ueno"
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Find the last name of the author with first name "Amal".
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CREATE TABLE authors (lname VARCHAR, fname VARCHAR)
|
SELECT lname FROM authors WHERE fname = "Amal"
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### Context: CREATE TABLE authors (lname VARCHAR, fname VARCHAR) ### Question: Find the last name of the author with first name "Amal". ### Answer: SELECT lname FROM authors WHERE fname = "Amal"
|
Find the first names of all the authors ordered in alphabetical order.
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CREATE TABLE authors (fname VARCHAR)
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SELECT fname FROM authors ORDER BY fname
|
### Context: CREATE TABLE authors (fname VARCHAR) ### Question: Find the first names of all the authors ordered in alphabetical order. ### Answer: SELECT fname FROM authors ORDER BY fname
|
Retrieve all the last names of authors in alphabetical order.
|
CREATE TABLE authors (lname VARCHAR)
|
SELECT lname FROM authors ORDER BY lname
|
### Context: CREATE TABLE authors (lname VARCHAR) ### Question: Retrieve all the last names of authors in alphabetical order. ### Answer: SELECT lname FROM authors ORDER BY lname
|
Retrieve all the first and last names of authors in the alphabetical order of last names.
|
CREATE TABLE authors (fname VARCHAR, lname VARCHAR)
|
SELECT fname, lname FROM authors ORDER BY lname
|
### Context: CREATE TABLE authors (fname VARCHAR, lname VARCHAR) ### Question: Retrieve all the first and last names of authors in the alphabetical order of last names. ### Answer: SELECT fname, lname FROM authors ORDER BY lname
|
How many different last names do the actors and actresses have?
|
CREATE TABLE actor (last_name VARCHAR)
|
SELECT COUNT(DISTINCT last_name) FROM actor
|
### Context: CREATE TABLE actor (last_name VARCHAR) ### Question: How many different last names do the actors and actresses have? ### Answer: SELECT COUNT(DISTINCT last_name) FROM actor
|
What is the most popular first name of the actors?
|
CREATE TABLE actor (first_name VARCHAR)
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SELECT first_name FROM actor GROUP BY first_name ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE actor (first_name VARCHAR) ### Question: What is the most popular first name of the actors? ### Answer: SELECT first_name FROM actor GROUP BY first_name ORDER BY COUNT(*) DESC LIMIT 1
|
What is the most popular full name of the actors?
|
CREATE TABLE actor (first_name VARCHAR, last_name VARCHAR)
|
SELECT first_name, last_name FROM actor GROUP BY first_name, last_name ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE actor (first_name VARCHAR, last_name VARCHAR) ### Question: What is the most popular full name of the actors? ### Answer: SELECT first_name, last_name FROM actor GROUP BY first_name, last_name ORDER BY COUNT(*) DESC LIMIT 1
|
Which districts have at least two addresses?
|
CREATE TABLE address (district VARCHAR)
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SELECT district FROM address GROUP BY district HAVING COUNT(*) >= 2
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### Context: CREATE TABLE address (district VARCHAR) ### Question: Which districts have at least two addresses? ### Answer: SELECT district FROM address GROUP BY district HAVING COUNT(*) >= 2
|
What is the phone number and postal code of the address 1031 Daugavpils Parkway?
|
CREATE TABLE address (phone VARCHAR, postal_code VARCHAR, address VARCHAR)
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SELECT phone, postal_code FROM address WHERE address = '1031 Daugavpils Parkway'
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### Context: CREATE TABLE address (phone VARCHAR, postal_code VARCHAR, address VARCHAR) ### Question: What is the phone number and postal code of the address 1031 Daugavpils Parkway? ### Answer: SELECT phone, postal_code FROM address WHERE address = '1031 Daugavpils Parkway'
|
Which city has the most addresses? List the city name, number of addresses, and city id.
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CREATE TABLE address (city_id VARCHAR); CREATE TABLE city (city VARCHAR, city_id VARCHAR)
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SELECT T2.city, COUNT(*), T1.city_id FROM address AS T1 JOIN city AS T2 ON T1.city_id = T2.city_id GROUP BY T1.city_id ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE address (city_id VARCHAR); CREATE TABLE city (city VARCHAR, city_id VARCHAR) ### Question: Which city has the most addresses? List the city name, number of addresses, and city id. ### Answer: SELECT T2.city, COUNT(*), T1.city_id FROM address AS T1 JOIN city AS T2 ON T1.city_id = T2.city_id GROUP BY T1.city_id ORDER BY COUNT(*) DESC LIMIT 1
|
How many addresses are in the district of California?
|
CREATE TABLE address (district VARCHAR)
|
SELECT COUNT(*) FROM address WHERE district = 'California'
|
### Context: CREATE TABLE address (district VARCHAR) ### Question: How many addresses are in the district of California? ### Answer: SELECT COUNT(*) FROM address WHERE district = 'California'
|
Which film is rented at a fee of 0.99 and has less than 3 in the inventory? List the film title and id.
|
CREATE TABLE film (title VARCHAR, film_id VARCHAR, rental_rate VARCHAR); CREATE TABLE inventory (film_id VARCHAR); CREATE TABLE film (title VARCHAR, film_id VARCHAR)
|
SELECT title, film_id FROM film WHERE rental_rate = 0.99 INTERSECT SELECT T1.title, T1.film_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id GROUP BY T1.film_id HAVING COUNT(*) < 3
|
### Context: CREATE TABLE film (title VARCHAR, film_id VARCHAR, rental_rate VARCHAR); CREATE TABLE inventory (film_id VARCHAR); CREATE TABLE film (title VARCHAR, film_id VARCHAR) ### Question: Which film is rented at a fee of 0.99 and has less than 3 in the inventory? List the film title and id. ### Answer: SELECT title, film_id FROM film WHERE rental_rate = 0.99 INTERSECT SELECT T1.title, T1.film_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id GROUP BY T1.film_id HAVING COUNT(*) < 3
|
How many cities are in Australia?
|
CREATE TABLE country (country_id VARCHAR, country VARCHAR); CREATE TABLE city (country_id VARCHAR)
|
SELECT COUNT(*) FROM city AS T1 JOIN country AS T2 ON T1.country_id = T2.country_id WHERE T2.country = 'Australia'
|
### Context: CREATE TABLE country (country_id VARCHAR, country VARCHAR); CREATE TABLE city (country_id VARCHAR) ### Question: How many cities are in Australia? ### Answer: SELECT COUNT(*) FROM city AS T1 JOIN country AS T2 ON T1.country_id = T2.country_id WHERE T2.country = 'Australia'
|
Which countries have at least 3 cities?
|
CREATE TABLE country (country VARCHAR, country_id VARCHAR); CREATE TABLE city (country_id VARCHAR)
|
SELECT T2.country FROM city AS T1 JOIN country AS T2 ON T1.country_id = T2.country_id GROUP BY T2.country_id HAVING COUNT(*) >= 3
|
### Context: CREATE TABLE country (country VARCHAR, country_id VARCHAR); CREATE TABLE city (country_id VARCHAR) ### Question: Which countries have at least 3 cities? ### Answer: SELECT T2.country FROM city AS T1 JOIN country AS T2 ON T1.country_id = T2.country_id GROUP BY T2.country_id HAVING COUNT(*) >= 3
|
Find all the payment dates for the payments with an amount larger than 10 and the payments handled by a staff person with the first name Elsa.
|
CREATE TABLE payment (payment_date VARCHAR, staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, first_name VARCHAR); CREATE TABLE payment (payment_date VARCHAR, amount INTEGER)
|
SELECT payment_date FROM payment WHERE amount > 10 UNION SELECT T1.payment_date FROM payment AS T1 JOIN staff AS T2 ON T1.staff_id = T2.staff_id WHERE T2.first_name = 'Elsa'
|
### Context: CREATE TABLE payment (payment_date VARCHAR, staff_id VARCHAR); CREATE TABLE staff (staff_id VARCHAR, first_name VARCHAR); CREATE TABLE payment (payment_date VARCHAR, amount INTEGER) ### Question: Find all the payment dates for the payments with an amount larger than 10 and the payments handled by a staff person with the first name Elsa. ### Answer: SELECT payment_date FROM payment WHERE amount > 10 UNION SELECT T1.payment_date FROM payment AS T1 JOIN staff AS T2 ON T1.staff_id = T2.staff_id WHERE T2.first_name = 'Elsa'
|
How many customers have an active value of 1?
|
CREATE TABLE customer (active VARCHAR)
|
SELECT COUNT(*) FROM customer WHERE active = '1'
|
### Context: CREATE TABLE customer (active VARCHAR) ### Question: How many customers have an active value of 1? ### Answer: SELECT COUNT(*) FROM customer WHERE active = '1'
|
Which film has the highest rental rate? And what is the rate?
|
CREATE TABLE film (title VARCHAR, rental_rate VARCHAR)
|
SELECT title, rental_rate FROM film ORDER BY rental_rate DESC LIMIT 1
|
### Context: CREATE TABLE film (title VARCHAR, rental_rate VARCHAR) ### Question: Which film has the highest rental rate? And what is the rate? ### Answer: SELECT title, rental_rate FROM film ORDER BY rental_rate DESC LIMIT 1
|
Which film has the most number of actors or actresses? List the film name, film id and description.
|
CREATE TABLE film_actor (film_id VARCHAR); CREATE TABLE film (title VARCHAR, film_id VARCHAR, description VARCHAR)
|
SELECT T2.title, T2.film_id, T2.description FROM film_actor AS T1 JOIN film AS T2 ON T1.film_id = T2.film_id GROUP BY T2.film_id ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE film_actor (film_id VARCHAR); CREATE TABLE film (title VARCHAR, film_id VARCHAR, description VARCHAR) ### Question: Which film has the most number of actors or actresses? List the film name, film id and description. ### Answer: SELECT T2.title, T2.film_id, T2.description FROM film_actor AS T1 JOIN film AS T2 ON T1.film_id = T2.film_id GROUP BY T2.film_id ORDER BY COUNT(*) DESC LIMIT 1
|
Which film actor (actress) starred the most films? List his or her first name, last name and actor id.
|
CREATE TABLE film_actor (actor_id VARCHAR); CREATE TABLE actor (first_name VARCHAR, last_name VARCHAR, actor_id VARCHAR)
|
SELECT T2.first_name, T2.last_name, T2.actor_id FROM film_actor AS T1 JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.actor_id ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE film_actor (actor_id VARCHAR); CREATE TABLE actor (first_name VARCHAR, last_name VARCHAR, actor_id VARCHAR) ### Question: Which film actor (actress) starred the most films? List his or her first name, last name and actor id. ### Answer: SELECT T2.first_name, T2.last_name, T2.actor_id FROM film_actor AS T1 JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.actor_id ORDER BY COUNT(*) DESC LIMIT 1
|
Which film actors (actresses) played a role in more than 30 films? List his or her first name and last name.
|
CREATE TABLE film_actor (actor_id VARCHAR); CREATE TABLE actor (first_name VARCHAR, last_name VARCHAR, actor_id VARCHAR)
|
SELECT T2.first_name, T2.last_name FROM film_actor AS T1 JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.actor_id HAVING COUNT(*) > 30
|
### Context: CREATE TABLE film_actor (actor_id VARCHAR); CREATE TABLE actor (first_name VARCHAR, last_name VARCHAR, actor_id VARCHAR) ### Question: Which film actors (actresses) played a role in more than 30 films? List his or her first name and last name. ### Answer: SELECT T2.first_name, T2.last_name FROM film_actor AS T1 JOIN actor AS T2 ON T1.actor_id = T2.actor_id GROUP BY T2.actor_id HAVING COUNT(*) > 30
|
Which store owns most items?
|
CREATE TABLE inventory (store_id VARCHAR)
|
SELECT store_id FROM inventory GROUP BY store_id ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE inventory (store_id VARCHAR) ### Question: Which store owns most items? ### Answer: SELECT store_id FROM inventory GROUP BY store_id ORDER BY COUNT(*) DESC LIMIT 1
|
What is the total amount of all payments?
|
CREATE TABLE payment (amount INTEGER)
|
SELECT SUM(amount) FROM payment
|
### Context: CREATE TABLE payment (amount INTEGER) ### Question: What is the total amount of all payments? ### Answer: SELECT SUM(amount) FROM payment
|
Which customer, who has made at least one payment, has spent the least money? List his or her first name, last name, and the id.
|
CREATE TABLE payment (customer_id VARCHAR); CREATE TABLE customer (first_name VARCHAR, last_name VARCHAR, customer_id VARCHAR)
|
SELECT T1.first_name, T1.last_name, T1.customer_id FROM customer AS T1 JOIN payment AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY SUM(amount) LIMIT 1
|
### Context: CREATE TABLE payment (customer_id VARCHAR); CREATE TABLE customer (first_name VARCHAR, last_name VARCHAR, customer_id VARCHAR) ### Question: Which customer, who has made at least one payment, has spent the least money? List his or her first name, last name, and the id. ### Answer: SELECT T1.first_name, T1.last_name, T1.customer_id FROM customer AS T1 JOIN payment AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY SUM(amount) LIMIT 1
|
What is the genre name of the film HUNGER ROOF?
|
CREATE TABLE film_category (category_id VARCHAR, film_id VARCHAR); CREATE TABLE film (film_id VARCHAR, title VARCHAR); CREATE TABLE category (name VARCHAR, category_id VARCHAR)
|
SELECT T1.name FROM category AS T1 JOIN film_category AS T2 ON T1.category_id = T2.category_id JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T3.title = 'HUNGER ROOF'
|
### Context: CREATE TABLE film_category (category_id VARCHAR, film_id VARCHAR); CREATE TABLE film (film_id VARCHAR, title VARCHAR); CREATE TABLE category (name VARCHAR, category_id VARCHAR) ### Question: What is the genre name of the film HUNGER ROOF? ### Answer: SELECT T1.name FROM category AS T1 JOIN film_category AS T2 ON T1.category_id = T2.category_id JOIN film AS T3 ON T2.film_id = T3.film_id WHERE T3.title = 'HUNGER ROOF'
|
How many films are there in each category? List the genre name, genre id and the count.
|
CREATE TABLE film_category (category_id VARCHAR); CREATE TABLE category (name VARCHAR, category_id VARCHAR)
|
SELECT T2.name, T1.category_id, COUNT(*) FROM film_category AS T1 JOIN category AS T2 ON T1.category_id = T2.category_id GROUP BY T1.category_id
|
### Context: CREATE TABLE film_category (category_id VARCHAR); CREATE TABLE category (name VARCHAR, category_id VARCHAR) ### Question: How many films are there in each category? List the genre name, genre id and the count. ### Answer: SELECT T2.name, T1.category_id, COUNT(*) FROM film_category AS T1 JOIN category AS T2 ON T1.category_id = T2.category_id GROUP BY T1.category_id
|
Which film has the most copies in the inventory? List both title and id.
|
CREATE TABLE film (title VARCHAR, film_id VARCHAR); CREATE TABLE inventory (film_id VARCHAR)
|
SELECT T1.title, T1.film_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id GROUP BY T1.film_id ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE film (title VARCHAR, film_id VARCHAR); CREATE TABLE inventory (film_id VARCHAR) ### Question: Which film has the most copies in the inventory? List both title and id. ### Answer: SELECT T1.title, T1.film_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id GROUP BY T1.film_id ORDER BY COUNT(*) DESC LIMIT 1
|
What is the film title and inventory id of the item in the inventory which was rented most frequently?
|
CREATE TABLE film (title VARCHAR, film_id VARCHAR); CREATE TABLE inventory (inventory_id VARCHAR, film_id VARCHAR); CREATE TABLE rental (inventory_id VARCHAR)
|
SELECT T1.title, T2.inventory_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id JOIN rental AS T3 ON T2.inventory_id = T3.inventory_id GROUP BY T2.inventory_id ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE film (title VARCHAR, film_id VARCHAR); CREATE TABLE inventory (inventory_id VARCHAR, film_id VARCHAR); CREATE TABLE rental (inventory_id VARCHAR) ### Question: What is the film title and inventory id of the item in the inventory which was rented most frequently? ### Answer: SELECT T1.title, T2.inventory_id FROM film AS T1 JOIN inventory AS T2 ON T1.film_id = T2.film_id JOIN rental AS T3 ON T2.inventory_id = T3.inventory_id GROUP BY T2.inventory_id ORDER BY COUNT(*) DESC LIMIT 1
|
How many languages are in these films?
|
CREATE TABLE film (language_id VARCHAR)
|
SELECT COUNT(DISTINCT language_id) FROM film
|
### Context: CREATE TABLE film (language_id VARCHAR) ### Question: How many languages are in these films? ### Answer: SELECT COUNT(DISTINCT language_id) FROM film
|
What are all the movies rated as R? List the titles.
|
CREATE TABLE film (title VARCHAR, rating VARCHAR)
|
SELECT title FROM film WHERE rating = 'R'
|
### Context: CREATE TABLE film (title VARCHAR, rating VARCHAR) ### Question: What are all the movies rated as R? List the titles. ### Answer: SELECT title FROM film WHERE rating = 'R'
|
Where is store 1 located?
|
CREATE TABLE store (address_id VARCHAR); CREATE TABLE address (address VARCHAR, address_id VARCHAR)
|
SELECT T2.address FROM store AS T1 JOIN address AS T2 ON T1.address_id = T2.address_id WHERE store_id = 1
|
### Context: CREATE TABLE store (address_id VARCHAR); CREATE TABLE address (address VARCHAR, address_id VARCHAR) ### Question: Where is store 1 located? ### Answer: SELECT T2.address FROM store AS T1 JOIN address AS T2 ON T1.address_id = T2.address_id WHERE store_id = 1
|
Which staff handled least number of payments? List the full name and the id.
|
CREATE TABLE payment (staff_id VARCHAR); CREATE TABLE staff (first_name VARCHAR, last_name VARCHAR, staff_id VARCHAR)
|
SELECT T1.first_name, T1.last_name, T1.staff_id FROM staff AS T1 JOIN payment AS T2 ON T1.staff_id = T2.staff_id GROUP BY T1.staff_id ORDER BY COUNT(*) LIMIT 1
|
### Context: CREATE TABLE payment (staff_id VARCHAR); CREATE TABLE staff (first_name VARCHAR, last_name VARCHAR, staff_id VARCHAR) ### Question: Which staff handled least number of payments? List the full name and the id. ### Answer: SELECT T1.first_name, T1.last_name, T1.staff_id FROM staff AS T1 JOIN payment AS T2 ON T1.staff_id = T2.staff_id GROUP BY T1.staff_id ORDER BY COUNT(*) LIMIT 1
|
Which language does the film AIRPORT POLLOCK use? List the language name.
|
CREATE TABLE film (language_id VARCHAR, title VARCHAR); CREATE TABLE LANGUAGE (name VARCHAR, language_id VARCHAR)
|
SELECT T2.name FROM film AS T1 JOIN LANGUAGE AS T2 ON T1.language_id = T2.language_id WHERE T1.title = 'AIRPORT POLLOCK'
|
### Context: CREATE TABLE film (language_id VARCHAR, title VARCHAR); CREATE TABLE LANGUAGE (name VARCHAR, language_id VARCHAR) ### Question: Which language does the film AIRPORT POLLOCK use? List the language name. ### Answer: SELECT T2.name FROM film AS T1 JOIN LANGUAGE AS T2 ON T1.language_id = T2.language_id WHERE T1.title = 'AIRPORT POLLOCK'
|
How many stores are there?
|
CREATE TABLE store (Id VARCHAR)
|
SELECT COUNT(*) FROM store
|
### Context: CREATE TABLE store (Id VARCHAR) ### Question: How many stores are there? ### Answer: SELECT COUNT(*) FROM store
|
How many kinds of different ratings are listed?
|
CREATE TABLE film (rating VARCHAR)
|
SELECT COUNT(DISTINCT rating) FROM film
|
### Context: CREATE TABLE film (rating VARCHAR) ### Question: How many kinds of different ratings are listed? ### Answer: SELECT COUNT(DISTINCT rating) FROM film
|
Which movies have 'Deleted Scenes' as a substring in the special feature?
|
CREATE TABLE film (title VARCHAR, special_features VARCHAR)
|
SELECT title FROM film WHERE special_features LIKE '%Deleted Scenes%'
|
### Context: CREATE TABLE film (title VARCHAR, special_features VARCHAR) ### Question: Which movies have 'Deleted Scenes' as a substring in the special feature? ### Answer: SELECT title FROM film WHERE special_features LIKE '%Deleted Scenes%'
|
How many items in inventory does store 1 have?
|
CREATE TABLE inventory (store_id VARCHAR)
|
SELECT COUNT(*) FROM inventory WHERE store_id = 1
|
### Context: CREATE TABLE inventory (store_id VARCHAR) ### Question: How many items in inventory does store 1 have? ### Answer: SELECT COUNT(*) FROM inventory WHERE store_id = 1
|
When did the first payment happen?
|
CREATE TABLE payment (payment_date VARCHAR)
|
SELECT payment_date FROM payment ORDER BY payment_date LIMIT 1
|
### Context: CREATE TABLE payment (payment_date VARCHAR) ### Question: When did the first payment happen? ### Answer: SELECT payment_date FROM payment ORDER BY payment_date LIMIT 1
|
Where does the customer with the first name Linda live? And what is her email?
|
CREATE TABLE customer (email VARCHAR, address_id VARCHAR, first_name VARCHAR); CREATE TABLE address (address VARCHAR, address_id VARCHAR)
|
SELECT T2.address, T1.email FROM customer AS T1 JOIN address AS T2 ON T2.address_id = T1.address_id WHERE T1.first_name = 'LINDA'
|
### Context: CREATE TABLE customer (email VARCHAR, address_id VARCHAR, first_name VARCHAR); CREATE TABLE address (address VARCHAR, address_id VARCHAR) ### Question: Where does the customer with the first name Linda live? And what is her email? ### Answer: SELECT T2.address, T1.email FROM customer AS T1 JOIN address AS T2 ON T2.address_id = T1.address_id WHERE T1.first_name = 'LINDA'
|
Find all the films longer than 100 minutes, or rated PG, except those who cost more than 200 for replacement. List the titles.
|
CREATE TABLE film (title VARCHAR, replacement_cost INTEGER, LENGTH VARCHAR, rating VARCHAR)
|
SELECT title FROM film WHERE LENGTH > 100 OR rating = 'PG' EXCEPT SELECT title FROM film WHERE replacement_cost > 200
|
### Context: CREATE TABLE film (title VARCHAR, replacement_cost INTEGER, LENGTH VARCHAR, rating VARCHAR) ### Question: Find all the films longer than 100 minutes, or rated PG, except those who cost more than 200 for replacement. List the titles. ### Answer: SELECT title FROM film WHERE LENGTH > 100 OR rating = 'PG' EXCEPT SELECT title FROM film WHERE replacement_cost > 200
|
What is the first name and the last name of the customer who made the earliest rental?
|
CREATE TABLE customer (first_name VARCHAR, last_name VARCHAR, customer_id VARCHAR); CREATE TABLE rental (customer_id VARCHAR, rental_date VARCHAR)
|
SELECT T1.first_name, T1.last_name FROM customer AS T1 JOIN rental AS T2 ON T1.customer_id = T2.customer_id ORDER BY T2.rental_date LIMIT 1
|
### Context: CREATE TABLE customer (first_name VARCHAR, last_name VARCHAR, customer_id VARCHAR); CREATE TABLE rental (customer_id VARCHAR, rental_date VARCHAR) ### Question: What is the first name and the last name of the customer who made the earliest rental? ### Answer: SELECT T1.first_name, T1.last_name FROM customer AS T1 JOIN rental AS T2 ON T1.customer_id = T2.customer_id ORDER BY T2.rental_date LIMIT 1
|
What is the full name of the staff member who has rented a film to a customer with the first name April and the last name Burns?
|
CREATE TABLE customer (customer_id VARCHAR, first_name VARCHAR, last_name VARCHAR); CREATE TABLE rental (staff_id VARCHAR, customer_id VARCHAR); CREATE TABLE staff (first_name VARCHAR, last_name VARCHAR, staff_id VARCHAR)
|
SELECT DISTINCT T1.first_name, T1.last_name FROM staff AS T1 JOIN rental AS T2 ON T1.staff_id = T2.staff_id JOIN customer AS T3 ON T2.customer_id = T3.customer_id WHERE T3.first_name = 'APRIL' AND T3.last_name = 'BURNS'
|
### Context: CREATE TABLE customer (customer_id VARCHAR, first_name VARCHAR, last_name VARCHAR); CREATE TABLE rental (staff_id VARCHAR, customer_id VARCHAR); CREATE TABLE staff (first_name VARCHAR, last_name VARCHAR, staff_id VARCHAR) ### Question: What is the full name of the staff member who has rented a film to a customer with the first name April and the last name Burns? ### Answer: SELECT DISTINCT T1.first_name, T1.last_name FROM staff AS T1 JOIN rental AS T2 ON T1.staff_id = T2.staff_id JOIN customer AS T3 ON T2.customer_id = T3.customer_id WHERE T3.first_name = 'APRIL' AND T3.last_name = 'BURNS'
|
Which store has most the customers?
|
CREATE TABLE customer (store_id VARCHAR)
|
SELECT store_id FROM customer GROUP BY store_id ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE customer (store_id VARCHAR) ### Question: Which store has most the customers? ### Answer: SELECT store_id FROM customer GROUP BY store_id ORDER BY COUNT(*) DESC LIMIT 1
|
What is the largest payment amount?
|
CREATE TABLE payment (amount VARCHAR)
|
SELECT amount FROM payment ORDER BY amount DESC LIMIT 1
|
### Context: CREATE TABLE payment (amount VARCHAR) ### Question: What is the largest payment amount? ### Answer: SELECT amount FROM payment ORDER BY amount DESC LIMIT 1
|
Where does the staff member with the first name Elsa live?
|
CREATE TABLE staff (address_id VARCHAR, first_name VARCHAR); CREATE TABLE address (address VARCHAR, address_id VARCHAR)
|
SELECT T2.address FROM staff AS T1 JOIN address AS T2 ON T1.address_id = T2.address_id WHERE T1.first_name = 'Elsa'
|
### Context: CREATE TABLE staff (address_id VARCHAR, first_name VARCHAR); CREATE TABLE address (address VARCHAR, address_id VARCHAR) ### Question: Where does the staff member with the first name Elsa live? ### Answer: SELECT T2.address FROM staff AS T1 JOIN address AS T2 ON T1.address_id = T2.address_id WHERE T1.first_name = 'Elsa'
|
What are the first names of customers who have not rented any films after '2005-08-23 02:06:01'?
|
CREATE TABLE customer (first_name VARCHAR, customer_id VARCHAR, rental_date INTEGER); CREATE TABLE rental (first_name VARCHAR, customer_id VARCHAR, rental_date INTEGER)
|
SELECT first_name FROM customer WHERE NOT customer_id IN (SELECT customer_id FROM rental WHERE rental_date > '2005-08-23 02:06:01')
|
### Context: CREATE TABLE customer (first_name VARCHAR, customer_id VARCHAR, rental_date INTEGER); CREATE TABLE rental (first_name VARCHAR, customer_id VARCHAR, rental_date INTEGER) ### Question: What are the first names of customers who have not rented any films after '2005-08-23 02:06:01'? ### Answer: SELECT first_name FROM customer WHERE NOT customer_id IN (SELECT customer_id FROM rental WHERE rental_date > '2005-08-23 02:06:01')
|
How many bank branches are there?
|
CREATE TABLE bank (Id VARCHAR)
|
SELECT COUNT(*) FROM bank
|
### Context: CREATE TABLE bank (Id VARCHAR) ### Question: How many bank branches are there? ### Answer: SELECT COUNT(*) FROM bank
|
How many customers are there?
|
CREATE TABLE bank (no_of_customers INTEGER)
|
SELECT SUM(no_of_customers) FROM bank
|
### Context: CREATE TABLE bank (no_of_customers INTEGER) ### Question: How many customers are there? ### Answer: SELECT SUM(no_of_customers) FROM bank
|
Find the number of customers in the banks at New York City.
|
CREATE TABLE bank (no_of_customers INTEGER, city VARCHAR)
|
SELECT SUM(no_of_customers) FROM bank WHERE city = 'New York City'
|
### Context: CREATE TABLE bank (no_of_customers INTEGER, city VARCHAR) ### Question: Find the number of customers in the banks at New York City. ### Answer: SELECT SUM(no_of_customers) FROM bank WHERE city = 'New York City'
|
Find the average number of customers in all banks of Utah state.
|
CREATE TABLE bank (no_of_customers INTEGER, state VARCHAR)
|
SELECT AVG(no_of_customers) FROM bank WHERE state = 'Utah'
|
### Context: CREATE TABLE bank (no_of_customers INTEGER, state VARCHAR) ### Question: Find the average number of customers in all banks of Utah state. ### Answer: SELECT AVG(no_of_customers) FROM bank WHERE state = 'Utah'
|
Find the average number of customers cross all banks.
|
CREATE TABLE bank (no_of_customers INTEGER)
|
SELECT AVG(no_of_customers) FROM bank
|
### Context: CREATE TABLE bank (no_of_customers INTEGER) ### Question: Find the average number of customers cross all banks. ### Answer: SELECT AVG(no_of_customers) FROM bank
|
Find the city and state of the bank branch named morningside.
|
CREATE TABLE bank (city VARCHAR, state VARCHAR, bname VARCHAR)
|
SELECT city, state FROM bank WHERE bname = 'morningside'
|
### Context: CREATE TABLE bank (city VARCHAR, state VARCHAR, bname VARCHAR) ### Question: Find the city and state of the bank branch named morningside. ### Answer: SELECT city, state FROM bank WHERE bname = 'morningside'
|
Find the branch names of banks in the New York state.
|
CREATE TABLE bank (bname VARCHAR, state VARCHAR)
|
SELECT bname FROM bank WHERE state = 'New York'
|
### Context: CREATE TABLE bank (bname VARCHAR, state VARCHAR) ### Question: Find the branch names of banks in the New York state. ### Answer: SELECT bname FROM bank WHERE state = 'New York'
|
List the name of all customers sorted by their account balance in ascending order.
|
CREATE TABLE customer (cust_name VARCHAR, acc_bal VARCHAR)
|
SELECT cust_name FROM customer ORDER BY acc_bal
|
### Context: CREATE TABLE customer (cust_name VARCHAR, acc_bal VARCHAR) ### Question: List the name of all customers sorted by their account balance in ascending order. ### Answer: SELECT cust_name FROM customer ORDER BY acc_bal
|
List the name of all different customers who have some loan sorted by their total loan amount.
|
CREATE TABLE loan (cust_id VARCHAR, amount INTEGER); CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR)
|
SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id GROUP BY T1.cust_name ORDER BY SUM(T2.amount)
|
### Context: CREATE TABLE loan (cust_id VARCHAR, amount INTEGER); CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR) ### Question: List the name of all different customers who have some loan sorted by their total loan amount. ### Answer: SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id GROUP BY T1.cust_name ORDER BY SUM(T2.amount)
|
Find the state, account type, and credit score of the customer whose number of loan is 0.
|
CREATE TABLE customer (state VARCHAR, acc_type VARCHAR, credit_score VARCHAR, no_of_loans VARCHAR)
|
SELECT state, acc_type, credit_score FROM customer WHERE no_of_loans = 0
|
### Context: CREATE TABLE customer (state VARCHAR, acc_type VARCHAR, credit_score VARCHAR, no_of_loans VARCHAR) ### Question: Find the state, account type, and credit score of the customer whose number of loan is 0. ### Answer: SELECT state, acc_type, credit_score FROM customer WHERE no_of_loans = 0
|
Find the number of different cities which banks are located at.
|
CREATE TABLE bank (city VARCHAR)
|
SELECT COUNT(DISTINCT city) FROM bank
|
### Context: CREATE TABLE bank (city VARCHAR) ### Question: Find the number of different cities which banks are located at. ### Answer: SELECT COUNT(DISTINCT city) FROM bank
|
Find the number of different states which banks are located at.
|
CREATE TABLE bank (state VARCHAR)
|
SELECT COUNT(DISTINCT state) FROM bank
|
### Context: CREATE TABLE bank (state VARCHAR) ### Question: Find the number of different states which banks are located at. ### Answer: SELECT COUNT(DISTINCT state) FROM bank
|
How many distinct types of accounts are there?
|
CREATE TABLE customer (acc_type VARCHAR)
|
SELECT COUNT(DISTINCT acc_type) FROM customer
|
### Context: CREATE TABLE customer (acc_type VARCHAR) ### Question: How many distinct types of accounts are there? ### Answer: SELECT COUNT(DISTINCT acc_type) FROM customer
|
Find the name and account balance of the customer whose name includes the letter ‘a’.
|
CREATE TABLE customer (cust_name VARCHAR, acc_bal VARCHAR)
|
SELECT cust_name, acc_bal FROM customer WHERE cust_name LIKE '%a%'
|
### Context: CREATE TABLE customer (cust_name VARCHAR, acc_bal VARCHAR) ### Question: Find the name and account balance of the customer whose name includes the letter ‘a’. ### Answer: SELECT cust_name, acc_bal FROM customer WHERE cust_name LIKE '%a%'
|
Find the total account balance of each customer from Utah or Texas.
|
CREATE TABLE customer (acc_bal INTEGER, state VARCHAR)
|
SELECT SUM(acc_bal) FROM customer WHERE state = 'Utah' OR state = 'Texas'
|
### Context: CREATE TABLE customer (acc_bal INTEGER, state VARCHAR) ### Question: Find the total account balance of each customer from Utah or Texas. ### Answer: SELECT SUM(acc_bal) FROM customer WHERE state = 'Utah' OR state = 'Texas'
|
Find the name of customers who have both saving and checking account types.
|
CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR)
|
SELECT cust_name FROM customer WHERE acc_type = 'saving' INTERSECT SELECT cust_name FROM customer WHERE acc_type = 'checking'
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### Context: CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR) ### Question: Find the name of customers who have both saving and checking account types. ### Answer: SELECT cust_name FROM customer WHERE acc_type = 'saving' INTERSECT SELECT cust_name FROM customer WHERE acc_type = 'checking'
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Find the name of customers who do not have an saving account.
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CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR)
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SELECT cust_name FROM customer EXCEPT SELECT cust_name FROM customer WHERE acc_type = 'saving'
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### Context: CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR) ### Question: Find the name of customers who do not have an saving account. ### Answer: SELECT cust_name FROM customer EXCEPT SELECT cust_name FROM customer WHERE acc_type = 'saving'
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Find the name of customers who do not have a loan with a type of Mortgages.
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CREATE TABLE loan (cust_id VARCHAR, loan_type VARCHAR); CREATE TABLE customer (cust_name VARCHAR); CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR)
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SELECT cust_name FROM customer EXCEPT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE T2.loan_type = 'Mortgages'
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### Context: CREATE TABLE loan (cust_id VARCHAR, loan_type VARCHAR); CREATE TABLE customer (cust_name VARCHAR); CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR) ### Question: Find the name of customers who do not have a loan with a type of Mortgages. ### Answer: SELECT cust_name FROM customer EXCEPT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE T2.loan_type = 'Mortgages'
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Find the name of customers who have loans of both Mortgages and Auto.
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CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR); CREATE TABLE loan (cust_id VARCHAR)
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SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE loan_type = 'Mortgages' INTERSECT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE loan_type = 'Auto'
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### Context: CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR); CREATE TABLE loan (cust_id VARCHAR) ### Question: Find the name of customers who have loans of both Mortgages and Auto. ### Answer: SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE loan_type = 'Mortgages' INTERSECT SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE loan_type = 'Auto'
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Find the name of customers whose credit score is below the average credit scores of all customers.
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CREATE TABLE customer (cust_name VARCHAR, credit_score INTEGER)
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SELECT cust_name FROM customer WHERE credit_score < (SELECT AVG(credit_score) FROM customer)
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### Context: CREATE TABLE customer (cust_name VARCHAR, credit_score INTEGER) ### Question: Find the name of customers whose credit score is below the average credit scores of all customers. ### Answer: SELECT cust_name FROM customer WHERE credit_score < (SELECT AVG(credit_score) FROM customer)
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Find the branch name of the bank that has the most number of customers.
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CREATE TABLE bank (bname VARCHAR, no_of_customers VARCHAR)
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SELECT bname FROM bank ORDER BY no_of_customers DESC LIMIT 1
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### Context: CREATE TABLE bank (bname VARCHAR, no_of_customers VARCHAR) ### Question: Find the branch name of the bank that has the most number of customers. ### Answer: SELECT bname FROM bank ORDER BY no_of_customers DESC LIMIT 1
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Find the name of customer who has the lowest credit score.
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CREATE TABLE customer (cust_name VARCHAR, credit_score VARCHAR)
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SELECT cust_name FROM customer ORDER BY credit_score LIMIT 1
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### Context: CREATE TABLE customer (cust_name VARCHAR, credit_score VARCHAR) ### Question: Find the name of customer who has the lowest credit score. ### Answer: SELECT cust_name FROM customer ORDER BY credit_score LIMIT 1
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Find the name, account type, and account balance of the customer who has the highest credit score.
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CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR, acc_bal VARCHAR, credit_score VARCHAR)
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SELECT cust_name, acc_type, acc_bal FROM customer ORDER BY credit_score DESC LIMIT 1
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### Context: CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR, acc_bal VARCHAR, credit_score VARCHAR) ### Question: Find the name, account type, and account balance of the customer who has the highest credit score. ### Answer: SELECT cust_name, acc_type, acc_bal FROM customer ORDER BY credit_score DESC LIMIT 1
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Find the name of customer who has the highest amount of loans.
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CREATE TABLE loan (cust_id VARCHAR, amount INTEGER); CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR)
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SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id GROUP BY T1.cust_name ORDER BY SUM(T2.amount) DESC LIMIT 1
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### Context: CREATE TABLE loan (cust_id VARCHAR, amount INTEGER); CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR) ### Question: Find the name of customer who has the highest amount of loans. ### Answer: SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id GROUP BY T1.cust_name ORDER BY SUM(T2.amount) DESC LIMIT 1
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Find the state which has the most number of customers.
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CREATE TABLE bank (state VARCHAR, no_of_customers INTEGER)
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SELECT state FROM bank GROUP BY state ORDER BY SUM(no_of_customers) DESC LIMIT 1
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### Context: CREATE TABLE bank (state VARCHAR, no_of_customers INTEGER) ### Question: Find the state which has the most number of customers. ### Answer: SELECT state FROM bank GROUP BY state ORDER BY SUM(no_of_customers) DESC LIMIT 1
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For each account type, find the average account balance of customers with credit score lower than 50.
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CREATE TABLE customer (acc_type VARCHAR, acc_bal INTEGER, credit_score INTEGER)
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SELECT AVG(acc_bal), acc_type FROM customer WHERE credit_score < 50 GROUP BY acc_type
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### Context: CREATE TABLE customer (acc_type VARCHAR, acc_bal INTEGER, credit_score INTEGER) ### Question: For each account type, find the average account balance of customers with credit score lower than 50. ### Answer: SELECT AVG(acc_bal), acc_type FROM customer WHERE credit_score < 50 GROUP BY acc_type
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For each state, find the total account balance of customers whose credit score is above 100.
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CREATE TABLE customer (state VARCHAR, acc_bal INTEGER, credit_score INTEGER)
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SELECT SUM(acc_bal), state FROM customer WHERE credit_score > 100 GROUP BY state
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### Context: CREATE TABLE customer (state VARCHAR, acc_bal INTEGER, credit_score INTEGER) ### Question: For each state, find the total account balance of customers whose credit score is above 100. ### Answer: SELECT SUM(acc_bal), state FROM customer WHERE credit_score > 100 GROUP BY state
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