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Find all the name of documents without any sections.
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CREATE TABLE document_sections (document_name VARCHAR, document_code VARCHAR); CREATE TABLE documents (document_name VARCHAR, document_code VARCHAR)
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SELECT document_name FROM documents WHERE NOT document_code IN (SELECT document_code FROM document_sections)
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### Context: CREATE TABLE document_sections (document_name VARCHAR, document_code VARCHAR); CREATE TABLE documents (document_name VARCHAR, document_code VARCHAR) ### Question: Find all the name of documents without any sections. ### Answer: SELECT document_name FROM documents WHERE NOT document_code IN (SELECT document_code FROM document_sections)
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List all the username and passwords of users with the most popular role.
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CREATE TABLE users (user_name VARCHAR, password VARCHAR, role_code VARCHAR)
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SELECT user_name, password FROM users GROUP BY role_code ORDER BY COUNT(*) DESC LIMIT 1
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### Context: CREATE TABLE users (user_name VARCHAR, password VARCHAR, role_code VARCHAR) ### Question: List all the username and passwords of users with the most popular role. ### Answer: SELECT user_name, password FROM users GROUP BY role_code ORDER BY COUNT(*) DESC LIMIT 1
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Find the average access counts of documents with functional area "Acknowledgement".
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CREATE TABLE document_functional_areas (document_code VARCHAR, functional_area_code VARCHAR); CREATE TABLE documents (access_count INTEGER, document_code VARCHAR); CREATE TABLE functional_areas (functional_area_code VARCHAR, functional_area_description VARCHAR)
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SELECT AVG(t1.access_count) FROM documents AS t1 JOIN document_functional_areas AS t2 ON t1.document_code = t2.document_code JOIN functional_areas AS t3 ON t2.functional_area_code = t3.functional_area_code WHERE t3.functional_area_description = "Acknowledgement"
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### Context: CREATE TABLE document_functional_areas (document_code VARCHAR, functional_area_code VARCHAR); CREATE TABLE documents (access_count INTEGER, document_code VARCHAR); CREATE TABLE functional_areas (functional_area_code VARCHAR, functional_area_description VARCHAR) ### Question: Find the average access counts of documents with functional area "Acknowledgement". ### Answer: SELECT AVG(t1.access_count) FROM documents AS t1 JOIN document_functional_areas AS t2 ON t1.document_code = t2.document_code JOIN functional_areas AS t3 ON t2.functional_area_code = t3.functional_area_code WHERE t3.functional_area_description = "Acknowledgement"
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Find names of the document without any images.
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CREATE TABLE document_sections_images (section_id VARCHAR); CREATE TABLE documents (document_name VARCHAR); CREATE TABLE documents (document_name VARCHAR, document_code VARCHAR); CREATE TABLE document_sections (document_code VARCHAR, section_id VARCHAR)
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SELECT document_name FROM documents EXCEPT SELECT t1.document_name FROM documents AS t1 JOIN document_sections AS t2 ON t1.document_code = t2.document_code JOIN document_sections_images AS t3 ON t2.section_id = t3.section_id
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### Context: CREATE TABLE document_sections_images (section_id VARCHAR); CREATE TABLE documents (document_name VARCHAR); CREATE TABLE documents (document_name VARCHAR, document_code VARCHAR); CREATE TABLE document_sections (document_code VARCHAR, section_id VARCHAR) ### Question: Find names of the document without any images. ### Answer: SELECT document_name FROM documents EXCEPT SELECT t1.document_name FROM documents AS t1 JOIN document_sections AS t2 ON t1.document_code = t2.document_code JOIN document_sections_images AS t3 ON t2.section_id = t3.section_id
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What is the name of the document with the most number of sections?
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CREATE TABLE document_sections (document_code VARCHAR); CREATE TABLE documents (document_name VARCHAR, document_code VARCHAR)
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SELECT t1.document_name FROM documents AS t1 JOIN document_sections AS t2 ON t1.document_code = t2.document_code GROUP BY t1.document_code ORDER BY COUNT(*) DESC LIMIT 1
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### Context: CREATE TABLE document_sections (document_code VARCHAR); CREATE TABLE documents (document_name VARCHAR, document_code VARCHAR) ### Question: What is the name of the document with the most number of sections? ### Answer: SELECT t1.document_name FROM documents AS t1 JOIN document_sections AS t2 ON t1.document_code = t2.document_code GROUP BY t1.document_code ORDER BY COUNT(*) DESC LIMIT 1
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List all the document names which contains "CV".
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CREATE TABLE documents (document_name VARCHAR)
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SELECT document_name FROM documents WHERE document_name LIKE "%CV%"
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### Context: CREATE TABLE documents (document_name VARCHAR) ### Question: List all the document names which contains "CV". ### Answer: SELECT document_name FROM documents WHERE document_name LIKE "%CV%"
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How many users are logged in?
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CREATE TABLE users (user_login VARCHAR)
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SELECT COUNT(*) FROM users WHERE user_login = 1
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### Context: CREATE TABLE users (user_login VARCHAR) ### Question: How many users are logged in? ### Answer: SELECT COUNT(*) FROM users WHERE user_login = 1
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Find the description of the most popular role among the users that have logged in.
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CREATE TABLE users (role_description VARCHAR, role_code VARCHAR, user_login VARCHAR); CREATE TABLE ROLES (role_description VARCHAR, role_code VARCHAR, user_login VARCHAR)
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SELECT role_description FROM ROLES WHERE role_code = (SELECT role_code FROM users WHERE user_login = 1 GROUP BY role_code ORDER BY COUNT(*) DESC LIMIT 1)
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### Context: CREATE TABLE users (role_description VARCHAR, role_code VARCHAR, user_login VARCHAR); CREATE TABLE ROLES (role_description VARCHAR, role_code VARCHAR, user_login VARCHAR) ### Question: Find the description of the most popular role among the users that have logged in. ### Answer: SELECT role_description FROM ROLES WHERE role_code = (SELECT role_code FROM users WHERE user_login = 1 GROUP BY role_code ORDER BY COUNT(*) DESC LIMIT 1)
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Find the average access count of documents with the least popular structure.
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CREATE TABLE documents (access_count INTEGER, document_structure_code VARCHAR)
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SELECT AVG(access_count) FROM documents GROUP BY document_structure_code ORDER BY COUNT(*) LIMIT 1
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### Context: CREATE TABLE documents (access_count INTEGER, document_structure_code VARCHAR) ### Question: Find the average access count of documents with the least popular structure. ### Answer: SELECT AVG(access_count) FROM documents GROUP BY document_structure_code ORDER BY COUNT(*) LIMIT 1
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List all the image name and URLs in the order of their names.
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CREATE TABLE images (image_name VARCHAR, image_url VARCHAR)
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SELECT image_name, image_url FROM images ORDER BY image_name
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### Context: CREATE TABLE images (image_name VARCHAR, image_url VARCHAR) ### Question: List all the image name and URLs in the order of their names. ### Answer: SELECT image_name, image_url FROM images ORDER BY image_name
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Find the number of users in each role.
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CREATE TABLE users (role_code VARCHAR)
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SELECT COUNT(*), role_code FROM users GROUP BY role_code
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### Context: CREATE TABLE users (role_code VARCHAR) ### Question: Find the number of users in each role. ### Answer: SELECT COUNT(*), role_code FROM users GROUP BY role_code
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What document types have more than 2 corresponding documents?
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CREATE TABLE documents (document_type_code VARCHAR)
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SELECT document_type_code FROM documents GROUP BY document_type_code HAVING COUNT(*) > 2
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### Context: CREATE TABLE documents (document_type_code VARCHAR) ### Question: What document types have more than 2 corresponding documents? ### Answer: SELECT document_type_code FROM documents GROUP BY document_type_code HAVING COUNT(*) > 2
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How many companies are there?
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CREATE TABLE Companies (Id VARCHAR)
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SELECT COUNT(*) FROM Companies
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### Context: CREATE TABLE Companies (Id VARCHAR) ### Question: How many companies are there? ### Answer: SELECT COUNT(*) FROM Companies
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List the names of companies in descending order of market value.
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CREATE TABLE Companies (name VARCHAR, Market_Value_billion VARCHAR)
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SELECT name FROM Companies ORDER BY Market_Value_billion DESC
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### Context: CREATE TABLE Companies (name VARCHAR, Market_Value_billion VARCHAR) ### Question: List the names of companies in descending order of market value. ### Answer: SELECT name FROM Companies ORDER BY Market_Value_billion DESC
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What are the names of companies whose headquarters are not "USA"?
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CREATE TABLE Companies (name VARCHAR, Headquarters VARCHAR)
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SELECT name FROM Companies WHERE Headquarters <> 'USA'
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### Context: CREATE TABLE Companies (name VARCHAR, Headquarters VARCHAR) ### Question: What are the names of companies whose headquarters are not "USA"? ### Answer: SELECT name FROM Companies WHERE Headquarters <> 'USA'
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What are the name and assets of each company, sorted in ascending order of company name?
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CREATE TABLE Companies (name VARCHAR, Assets_billion VARCHAR)
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SELECT name, Assets_billion FROM Companies ORDER BY name
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### Context: CREATE TABLE Companies (name VARCHAR, Assets_billion VARCHAR) ### Question: What are the name and assets of each company, sorted in ascending order of company name? ### Answer: SELECT name, Assets_billion FROM Companies ORDER BY name
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What are the average profits of companies?
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CREATE TABLE Companies (Profits_billion INTEGER)
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SELECT AVG(Profits_billion) FROM Companies
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### Context: CREATE TABLE Companies (Profits_billion INTEGER) ### Question: What are the average profits of companies? ### Answer: SELECT AVG(Profits_billion) FROM Companies
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What are the maximum and minimum sales of the companies whose industries are not "Banking".
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CREATE TABLE Companies (Sales_billion INTEGER, Industry VARCHAR)
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SELECT MAX(Sales_billion), MIN(Sales_billion) FROM Companies WHERE Industry <> "Banking"
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### Context: CREATE TABLE Companies (Sales_billion INTEGER, Industry VARCHAR) ### Question: What are the maximum and minimum sales of the companies whose industries are not "Banking". ### Answer: SELECT MAX(Sales_billion), MIN(Sales_billion) FROM Companies WHERE Industry <> "Banking"
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How many different industries are the companies in?
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CREATE TABLE Companies (Industry VARCHAR)
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SELECT COUNT(DISTINCT Industry) FROM Companies
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### Context: CREATE TABLE Companies (Industry VARCHAR) ### Question: How many different industries are the companies in? ### Answer: SELECT COUNT(DISTINCT Industry) FROM Companies
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List the names of buildings in descending order of building height.
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CREATE TABLE buildings (name VARCHAR, Height VARCHAR)
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SELECT name FROM buildings ORDER BY Height DESC
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### Context: CREATE TABLE buildings (name VARCHAR, Height VARCHAR) ### Question: List the names of buildings in descending order of building height. ### Answer: SELECT name FROM buildings ORDER BY Height DESC
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Find the stories of the building with the largest height.
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CREATE TABLE buildings (Stories VARCHAR, Height VARCHAR)
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SELECT Stories FROM buildings ORDER BY Height DESC LIMIT 1
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### Context: CREATE TABLE buildings (Stories VARCHAR, Height VARCHAR) ### Question: Find the stories of the building with the largest height. ### Answer: SELECT Stories FROM buildings ORDER BY Height DESC LIMIT 1
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List the name of a building along with the name of a company whose office is in the building.
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CREATE TABLE buildings (name VARCHAR, id VARCHAR); CREATE TABLE Office_locations (building_id VARCHAR, company_id VARCHAR); CREATE TABLE Companies (name VARCHAR, id VARCHAR)
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SELECT T3.name, T2.name FROM Office_locations AS T1 JOIN buildings AS T2 ON T1.building_id = T2.id JOIN Companies AS T3 ON T1.company_id = T3.id
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### Context: CREATE TABLE buildings (name VARCHAR, id VARCHAR); CREATE TABLE Office_locations (building_id VARCHAR, company_id VARCHAR); CREATE TABLE Companies (name VARCHAR, id VARCHAR) ### Question: List the name of a building along with the name of a company whose office is in the building. ### Answer: SELECT T3.name, T2.name FROM Office_locations AS T1 JOIN buildings AS T2 ON T1.building_id = T2.id JOIN Companies AS T3 ON T1.company_id = T3.id
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Show the names of the buildings that have more than one company offices.
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CREATE TABLE buildings (name VARCHAR, id VARCHAR); CREATE TABLE Companies (id VARCHAR); CREATE TABLE Office_locations (building_id VARCHAR, company_id VARCHAR)
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SELECT T2.name FROM Office_locations AS T1 JOIN buildings AS T2 ON T1.building_id = T2.id JOIN Companies AS T3 ON T1.company_id = T3.id GROUP BY T1.building_id HAVING COUNT(*) > 1
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### Context: CREATE TABLE buildings (name VARCHAR, id VARCHAR); CREATE TABLE Companies (id VARCHAR); CREATE TABLE Office_locations (building_id VARCHAR, company_id VARCHAR) ### Question: Show the names of the buildings that have more than one company offices. ### Answer: SELECT T2.name FROM Office_locations AS T1 JOIN buildings AS T2 ON T1.building_id = T2.id JOIN Companies AS T3 ON T1.company_id = T3.id GROUP BY T1.building_id HAVING COUNT(*) > 1
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Show the name of the building that has the most company offices.
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CREATE TABLE buildings (name VARCHAR, id VARCHAR); CREATE TABLE Companies (id VARCHAR); CREATE TABLE Office_locations (building_id VARCHAR, company_id VARCHAR)
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SELECT T2.name FROM Office_locations AS T1 JOIN buildings AS T2 ON T1.building_id = T2.id JOIN Companies AS T3 ON T1.company_id = T3.id GROUP BY T1.building_id ORDER BY COUNT(*) DESC LIMIT 1
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### Context: CREATE TABLE buildings (name VARCHAR, id VARCHAR); CREATE TABLE Companies (id VARCHAR); CREATE TABLE Office_locations (building_id VARCHAR, company_id VARCHAR) ### Question: Show the name of the building that has the most company offices. ### Answer: SELECT T2.name FROM Office_locations AS T1 JOIN buildings AS T2 ON T1.building_id = T2.id JOIN Companies AS T3 ON T1.company_id = T3.id GROUP BY T1.building_id ORDER BY COUNT(*) DESC LIMIT 1
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Please show the names of the buildings whose status is "on-hold", in ascending order of stories.
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CREATE TABLE buildings (name VARCHAR, Status VARCHAR, Stories VARCHAR)
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SELECT name FROM buildings WHERE Status = "on-hold" ORDER BY Stories
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### Context: CREATE TABLE buildings (name VARCHAR, Status VARCHAR, Stories VARCHAR) ### Question: Please show the names of the buildings whose status is "on-hold", in ascending order of stories. ### Answer: SELECT name FROM buildings WHERE Status = "on-hold" ORDER BY Stories
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Please show each industry and the corresponding number of companies in that industry.
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CREATE TABLE Companies (Industry VARCHAR)
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SELECT Industry, COUNT(*) FROM Companies GROUP BY Industry
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### Context: CREATE TABLE Companies (Industry VARCHAR) ### Question: Please show each industry and the corresponding number of companies in that industry. ### Answer: SELECT Industry, COUNT(*) FROM Companies GROUP BY Industry
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Please show the industries of companies in descending order of the number of companies.
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CREATE TABLE Companies (Industry VARCHAR)
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SELECT Industry FROM Companies GROUP BY Industry ORDER BY COUNT(*) DESC
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### Context: CREATE TABLE Companies (Industry VARCHAR) ### Question: Please show the industries of companies in descending order of the number of companies. ### Answer: SELECT Industry FROM Companies GROUP BY Industry ORDER BY COUNT(*) DESC
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List the industry shared by the most companies.
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CREATE TABLE Companies (Industry VARCHAR)
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SELECT Industry FROM Companies GROUP BY Industry ORDER BY COUNT(*) DESC LIMIT 1
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### Context: CREATE TABLE Companies (Industry VARCHAR) ### Question: List the industry shared by the most companies. ### Answer: SELECT Industry FROM Companies GROUP BY Industry ORDER BY COUNT(*) DESC LIMIT 1
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List the names of buildings that have no company office.
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CREATE TABLE buildings (name VARCHAR, id VARCHAR, building_id VARCHAR); CREATE TABLE Office_locations (name VARCHAR, id VARCHAR, building_id VARCHAR)
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SELECT name FROM buildings WHERE NOT id IN (SELECT building_id FROM Office_locations)
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### Context: CREATE TABLE buildings (name VARCHAR, id VARCHAR, building_id VARCHAR); CREATE TABLE Office_locations (name VARCHAR, id VARCHAR, building_id VARCHAR) ### Question: List the names of buildings that have no company office. ### Answer: SELECT name FROM buildings WHERE NOT id IN (SELECT building_id FROM Office_locations)
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Show the industries shared by companies whose headquarters are "USA" and companies whose headquarters are "China".
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CREATE TABLE Companies (Industry VARCHAR, Headquarters VARCHAR)
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SELECT Industry FROM Companies WHERE Headquarters = "USA" INTERSECT SELECT Industry FROM Companies WHERE Headquarters = "China"
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### Context: CREATE TABLE Companies (Industry VARCHAR, Headquarters VARCHAR) ### Question: Show the industries shared by companies whose headquarters are "USA" and companies whose headquarters are "China". ### Answer: SELECT Industry FROM Companies WHERE Headquarters = "USA" INTERSECT SELECT Industry FROM Companies WHERE Headquarters = "China"
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Find the number of companies whose industry is "Banking" or "Conglomerate",
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CREATE TABLE Companies (Industry VARCHAR)
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SELECT COUNT(*) FROM Companies WHERE Industry = "Banking" OR Industry = "Conglomerate"
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### Context: CREATE TABLE Companies (Industry VARCHAR) ### Question: Find the number of companies whose industry is "Banking" or "Conglomerate", ### Answer: SELECT COUNT(*) FROM Companies WHERE Industry = "Banking" OR Industry = "Conglomerate"
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Show the headquarters shared by more than two companies.
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CREATE TABLE Companies (Headquarters VARCHAR)
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SELECT Headquarters FROM Companies GROUP BY Headquarters HAVING COUNT(*) > 2
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### Context: CREATE TABLE Companies (Headquarters VARCHAR) ### Question: Show the headquarters shared by more than two companies. ### Answer: SELECT Headquarters FROM Companies GROUP BY Headquarters HAVING COUNT(*) > 2
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How many products are there?
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CREATE TABLE Products (Id VARCHAR)
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SELECT COUNT(*) FROM Products
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### Context: CREATE TABLE Products (Id VARCHAR) ### Question: How many products are there? ### Answer: SELECT COUNT(*) FROM Products
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List the name of products in ascending order of price.
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CREATE TABLE Products (Product_Name VARCHAR, Product_Price VARCHAR)
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SELECT Product_Name FROM Products ORDER BY Product_Price
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### Context: CREATE TABLE Products (Product_Name VARCHAR, Product_Price VARCHAR) ### Question: List the name of products in ascending order of price. ### Answer: SELECT Product_Name FROM Products ORDER BY Product_Price
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What are the names and type codes of products?
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CREATE TABLE Products (Product_Name VARCHAR, Product_Type_Code VARCHAR)
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SELECT Product_Name, Product_Type_Code FROM Products
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### Context: CREATE TABLE Products (Product_Name VARCHAR, Product_Type_Code VARCHAR) ### Question: What are the names and type codes of products? ### Answer: SELECT Product_Name, Product_Type_Code FROM Products
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Show the prices of the products named "Dining" or "Trading Policy".
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CREATE TABLE Products (Product_Price VARCHAR, Product_Name VARCHAR)
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SELECT Product_Price FROM Products WHERE Product_Name = "Dining" OR Product_Name = "Trading Policy"
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### Context: CREATE TABLE Products (Product_Price VARCHAR, Product_Name VARCHAR) ### Question: Show the prices of the products named "Dining" or "Trading Policy". ### Answer: SELECT Product_Price FROM Products WHERE Product_Name = "Dining" OR Product_Name = "Trading Policy"
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What is the average price for products?
|
CREATE TABLE Products (Product_Price INTEGER)
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SELECT AVG(Product_Price) FROM Products
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### Context: CREATE TABLE Products (Product_Price INTEGER) ### Question: What is the average price for products? ### Answer: SELECT AVG(Product_Price) FROM Products
|
What is the name of the product with the highest price?
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CREATE TABLE Products (Product_Name VARCHAR, Product_Price VARCHAR)
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SELECT Product_Name FROM Products ORDER BY Product_Price DESC LIMIT 1
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### Context: CREATE TABLE Products (Product_Name VARCHAR, Product_Price VARCHAR) ### Question: What is the name of the product with the highest price? ### Answer: SELECT Product_Name FROM Products ORDER BY Product_Price DESC LIMIT 1
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Show different type codes of products and the number of products with each type code.
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CREATE TABLE Products (Product_Type_Code VARCHAR)
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SELECT Product_Type_Code, COUNT(*) FROM Products GROUP BY Product_Type_Code
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### Context: CREATE TABLE Products (Product_Type_Code VARCHAR) ### Question: Show different type codes of products and the number of products with each type code. ### Answer: SELECT Product_Type_Code, COUNT(*) FROM Products GROUP BY Product_Type_Code
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Show the most common type code across products.
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CREATE TABLE Products (Product_Type_Code VARCHAR)
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SELECT Product_Type_Code FROM Products GROUP BY Product_Type_Code ORDER BY COUNT(*) DESC LIMIT 1
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### Context: CREATE TABLE Products (Product_Type_Code VARCHAR) ### Question: Show the most common type code across products. ### Answer: SELECT Product_Type_Code FROM Products GROUP BY Product_Type_Code ORDER BY COUNT(*) DESC LIMIT 1
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Show the product type codes that have at least two products.
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CREATE TABLE Products (Product_Type_Code VARCHAR)
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SELECT Product_Type_Code FROM Products GROUP BY Product_Type_Code HAVING COUNT(*) >= 2
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### Context: CREATE TABLE Products (Product_Type_Code VARCHAR) ### Question: Show the product type codes that have at least two products. ### Answer: SELECT Product_Type_Code FROM Products GROUP BY Product_Type_Code HAVING COUNT(*) >= 2
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Show the product type codes that have both products with price higher than 4500 and products with price lower than 3000.
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CREATE TABLE Products (Product_Type_Code VARCHAR, Product_Price INTEGER)
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SELECT Product_Type_Code FROM Products WHERE Product_Price > 4500 INTERSECT SELECT Product_Type_Code FROM Products WHERE Product_Price < 3000
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### Context: CREATE TABLE Products (Product_Type_Code VARCHAR, Product_Price INTEGER) ### Question: Show the product type codes that have both products with price higher than 4500 and products with price lower than 3000. ### Answer: SELECT Product_Type_Code FROM Products WHERE Product_Price > 4500 INTERSECT SELECT Product_Type_Code FROM Products WHERE Product_Price < 3000
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Show the names of products and the number of events they are in.
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CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_ID VARCHAR)
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SELECT T1.Product_Name, COUNT(*) FROM Products AS T1 JOIN Products_in_Events AS T2 ON T1.Product_ID = T2.Product_ID GROUP BY T1.Product_Name
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### Context: CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_ID VARCHAR) ### Question: Show the names of products and the number of events they are in. ### Answer: SELECT T1.Product_Name, COUNT(*) FROM Products AS T1 JOIN Products_in_Events AS T2 ON T1.Product_ID = T2.Product_ID GROUP BY T1.Product_Name
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Show the names of products and the number of events they are in, sorted by the number of events in descending order.
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CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_ID VARCHAR)
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SELECT T1.Product_Name, COUNT(*) FROM Products AS T1 JOIN Products_in_Events AS T2 ON T1.Product_ID = T2.Product_ID GROUP BY T1.Product_Name ORDER BY COUNT(*) DESC
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### Context: CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_ID VARCHAR) ### Question: Show the names of products and the number of events they are in, sorted by the number of events in descending order. ### Answer: SELECT T1.Product_Name, COUNT(*) FROM Products AS T1 JOIN Products_in_Events AS T2 ON T1.Product_ID = T2.Product_ID GROUP BY T1.Product_Name ORDER BY COUNT(*) DESC
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Show the names of products that are in at least two events.
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CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_ID VARCHAR)
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SELECT T1.Product_Name FROM Products AS T1 JOIN Products_in_Events AS T2 ON T1.Product_ID = T2.Product_ID GROUP BY T1.Product_Name HAVING COUNT(*) >= 2
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### Context: CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_ID VARCHAR) ### Question: Show the names of products that are in at least two events. ### Answer: SELECT T1.Product_Name FROM Products AS T1 JOIN Products_in_Events AS T2 ON T1.Product_ID = T2.Product_ID GROUP BY T1.Product_Name HAVING COUNT(*) >= 2
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Show the names of products that are in at least two events in ascending alphabetical order of product name.
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CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_ID VARCHAR)
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SELECT T1.Product_Name FROM Products AS T1 JOIN Products_in_Events AS T2 ON T1.Product_ID = T2.Product_ID GROUP BY T1.Product_Name HAVING COUNT(*) >= 2 ORDER BY T1.Product_Name
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### Context: CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_ID VARCHAR) ### Question: Show the names of products that are in at least two events in ascending alphabetical order of product name. ### Answer: SELECT T1.Product_Name FROM Products AS T1 JOIN Products_in_Events AS T2 ON T1.Product_ID = T2.Product_ID GROUP BY T1.Product_Name HAVING COUNT(*) >= 2 ORDER BY T1.Product_Name
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List the names of products that are not in any event.
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CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_Name VARCHAR, Product_ID VARCHAR)
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SELECT Product_Name FROM Products WHERE NOT Product_ID IN (SELECT Product_ID FROM Products_in_Events)
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### Context: CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_Name VARCHAR, Product_ID VARCHAR) ### Question: List the names of products that are not in any event. ### Answer: SELECT Product_Name FROM Products WHERE NOT Product_ID IN (SELECT Product_ID FROM Products_in_Events)
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How many artworks are there?
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CREATE TABLE artwork (Id VARCHAR)
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SELECT COUNT(*) FROM artwork
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### Context: CREATE TABLE artwork (Id VARCHAR) ### Question: How many artworks are there? ### Answer: SELECT COUNT(*) FROM artwork
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List the name of artworks in ascending alphabetical order.
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CREATE TABLE artwork (Name VARCHAR)
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SELECT Name FROM artwork ORDER BY Name
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### Context: CREATE TABLE artwork (Name VARCHAR) ### Question: List the name of artworks in ascending alphabetical order. ### Answer: SELECT Name FROM artwork ORDER BY Name
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List the name of artworks whose type is not "Program Talent Show".
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CREATE TABLE artwork (Name VARCHAR, TYPE VARCHAR)
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SELECT Name FROM artwork WHERE TYPE <> "Program Talent Show"
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### Context: CREATE TABLE artwork (Name VARCHAR, TYPE VARCHAR) ### Question: List the name of artworks whose type is not "Program Talent Show". ### Answer: SELECT Name FROM artwork WHERE TYPE <> "Program Talent Show"
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What are the names and locations of festivals?
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CREATE TABLE festival_detail (Festival_Name VARCHAR, LOCATION VARCHAR)
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SELECT Festival_Name, LOCATION FROM festival_detail
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### Context: CREATE TABLE festival_detail (Festival_Name VARCHAR, LOCATION VARCHAR) ### Question: What are the names and locations of festivals? ### Answer: SELECT Festival_Name, LOCATION FROM festival_detail
|
What are the names of the chairs of festivals, sorted in ascending order of the year held?
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CREATE TABLE festival_detail (Chair_Name VARCHAR, YEAR VARCHAR)
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SELECT Chair_Name FROM festival_detail ORDER BY YEAR
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### Context: CREATE TABLE festival_detail (Chair_Name VARCHAR, YEAR VARCHAR) ### Question: What are the names of the chairs of festivals, sorted in ascending order of the year held? ### Answer: SELECT Chair_Name FROM festival_detail ORDER BY YEAR
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What is the location of the festival with the largest number of audience?
|
CREATE TABLE festival_detail (LOCATION VARCHAR, Num_of_Audience VARCHAR)
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SELECT LOCATION FROM festival_detail ORDER BY Num_of_Audience DESC LIMIT 1
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### Context: CREATE TABLE festival_detail (LOCATION VARCHAR, Num_of_Audience VARCHAR) ### Question: What is the location of the festival with the largest number of audience? ### Answer: SELECT LOCATION FROM festival_detail ORDER BY Num_of_Audience DESC LIMIT 1
|
What are the names of festivals held in year 2007?
|
CREATE TABLE festival_detail (Festival_Name VARCHAR, YEAR VARCHAR)
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SELECT Festival_Name FROM festival_detail WHERE YEAR = 2007
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### Context: CREATE TABLE festival_detail (Festival_Name VARCHAR, YEAR VARCHAR) ### Question: What are the names of festivals held in year 2007? ### Answer: SELECT Festival_Name FROM festival_detail WHERE YEAR = 2007
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What is the average number of audience for festivals?
|
CREATE TABLE festival_detail (Num_of_Audience INTEGER)
|
SELECT AVG(Num_of_Audience) FROM festival_detail
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### Context: CREATE TABLE festival_detail (Num_of_Audience INTEGER) ### Question: What is the average number of audience for festivals? ### Answer: SELECT AVG(Num_of_Audience) FROM festival_detail
|
Show the names of the three most recent festivals.
|
CREATE TABLE festival_detail (Festival_Name VARCHAR, YEAR VARCHAR)
|
SELECT Festival_Name FROM festival_detail ORDER BY YEAR DESC LIMIT 3
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### Context: CREATE TABLE festival_detail (Festival_Name VARCHAR, YEAR VARCHAR) ### Question: Show the names of the three most recent festivals. ### Answer: SELECT Festival_Name FROM festival_detail ORDER BY YEAR DESC LIMIT 3
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For each nomination, show the name of the artwork and name of the festival where it is nominated.
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CREATE TABLE artwork (Name VARCHAR, Artwork_ID VARCHAR); CREATE TABLE nomination (Artwork_ID VARCHAR, Festival_ID VARCHAR); CREATE TABLE festival_detail (Festival_Name VARCHAR, Festival_ID VARCHAR)
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SELECT T2.Name, T3.Festival_Name FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID
|
### Context: CREATE TABLE artwork (Name VARCHAR, Artwork_ID VARCHAR); CREATE TABLE nomination (Artwork_ID VARCHAR, Festival_ID VARCHAR); CREATE TABLE festival_detail (Festival_Name VARCHAR, Festival_ID VARCHAR) ### Question: For each nomination, show the name of the artwork and name of the festival where it is nominated. ### Answer: SELECT T2.Name, T3.Festival_Name FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID
|
Show distinct types of artworks that are nominated in festivals in 2007.
|
CREATE TABLE nomination (Artwork_ID VARCHAR, Festival_ID VARCHAR); CREATE TABLE festival_detail (Festival_ID VARCHAR, Year VARCHAR); CREATE TABLE artwork (Type VARCHAR, Artwork_ID VARCHAR)
|
SELECT DISTINCT T2.Type FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID WHERE T3.Year = 2007
|
### Context: CREATE TABLE nomination (Artwork_ID VARCHAR, Festival_ID VARCHAR); CREATE TABLE festival_detail (Festival_ID VARCHAR, Year VARCHAR); CREATE TABLE artwork (Type VARCHAR, Artwork_ID VARCHAR) ### Question: Show distinct types of artworks that are nominated in festivals in 2007. ### Answer: SELECT DISTINCT T2.Type FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID WHERE T3.Year = 2007
|
Show the names of artworks in ascending order of the year they are nominated in.
|
CREATE TABLE artwork (Name VARCHAR, Artwork_ID VARCHAR); CREATE TABLE nomination (Artwork_ID VARCHAR, Festival_ID VARCHAR); CREATE TABLE festival_detail (Festival_ID VARCHAR, Year VARCHAR)
|
SELECT T2.Name FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID ORDER BY T3.Year
|
### Context: CREATE TABLE artwork (Name VARCHAR, Artwork_ID VARCHAR); CREATE TABLE nomination (Artwork_ID VARCHAR, Festival_ID VARCHAR); CREATE TABLE festival_detail (Festival_ID VARCHAR, Year VARCHAR) ### Question: Show the names of artworks in ascending order of the year they are nominated in. ### Answer: SELECT T2.Name FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID ORDER BY T3.Year
|
Show the names of festivals that have nominated artworks of type "Program Talent Show".
|
CREATE TABLE nomination (Artwork_ID VARCHAR, Festival_ID VARCHAR); CREATE TABLE artwork (Artwork_ID VARCHAR, Type VARCHAR); CREATE TABLE festival_detail (Festival_Name VARCHAR, Festival_ID VARCHAR)
|
SELECT T3.Festival_Name FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID WHERE T2.Type = "Program Talent Show"
|
### Context: CREATE TABLE nomination (Artwork_ID VARCHAR, Festival_ID VARCHAR); CREATE TABLE artwork (Artwork_ID VARCHAR, Type VARCHAR); CREATE TABLE festival_detail (Festival_Name VARCHAR, Festival_ID VARCHAR) ### Question: Show the names of festivals that have nominated artworks of type "Program Talent Show". ### Answer: SELECT T3.Festival_Name FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID WHERE T2.Type = "Program Talent Show"
|
Show the ids and names of festivals that have at least two nominations for artworks.
|
CREATE TABLE nomination (Festival_ID VARCHAR, Artwork_ID VARCHAR); CREATE TABLE festival_detail (Festival_Name VARCHAR, Festival_ID VARCHAR); CREATE TABLE artwork (Artwork_ID VARCHAR)
|
SELECT T1.Festival_ID, T3.Festival_Name FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID GROUP BY T1.Festival_ID HAVING COUNT(*) >= 2
|
### Context: CREATE TABLE nomination (Festival_ID VARCHAR, Artwork_ID VARCHAR); CREATE TABLE festival_detail (Festival_Name VARCHAR, Festival_ID VARCHAR); CREATE TABLE artwork (Artwork_ID VARCHAR) ### Question: Show the ids and names of festivals that have at least two nominations for artworks. ### Answer: SELECT T1.Festival_ID, T3.Festival_Name FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID GROUP BY T1.Festival_ID HAVING COUNT(*) >= 2
|
Show the id, name of each festival and the number of artworks it has nominated.
|
CREATE TABLE nomination (Festival_ID VARCHAR, Artwork_ID VARCHAR); CREATE TABLE festival_detail (Festival_Name VARCHAR, Festival_ID VARCHAR); CREATE TABLE artwork (Artwork_ID VARCHAR)
|
SELECT T1.Festival_ID, T3.Festival_Name, COUNT(*) FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID GROUP BY T1.Festival_ID
|
### Context: CREATE TABLE nomination (Festival_ID VARCHAR, Artwork_ID VARCHAR); CREATE TABLE festival_detail (Festival_Name VARCHAR, Festival_ID VARCHAR); CREATE TABLE artwork (Artwork_ID VARCHAR) ### Question: Show the id, name of each festival and the number of artworks it has nominated. ### Answer: SELECT T1.Festival_ID, T3.Festival_Name, COUNT(*) FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID GROUP BY T1.Festival_ID
|
Please show different types of artworks with the corresponding number of artworks of each type.
|
CREATE TABLE artwork (TYPE VARCHAR)
|
SELECT TYPE, COUNT(*) FROM artwork GROUP BY TYPE
|
### Context: CREATE TABLE artwork (TYPE VARCHAR) ### Question: Please show different types of artworks with the corresponding number of artworks of each type. ### Answer: SELECT TYPE, COUNT(*) FROM artwork GROUP BY TYPE
|
List the most common type of artworks.
|
CREATE TABLE artwork (TYPE VARCHAR)
|
SELECT TYPE FROM artwork GROUP BY TYPE ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE artwork (TYPE VARCHAR) ### Question: List the most common type of artworks. ### Answer: SELECT TYPE FROM artwork GROUP BY TYPE ORDER BY COUNT(*) DESC LIMIT 1
|
List the year in which there are more than one festivals.
|
CREATE TABLE festival_detail (YEAR VARCHAR)
|
SELECT YEAR FROM festival_detail GROUP BY YEAR HAVING COUNT(*) > 1
|
### Context: CREATE TABLE festival_detail (YEAR VARCHAR) ### Question: List the year in which there are more than one festivals. ### Answer: SELECT YEAR FROM festival_detail GROUP BY YEAR HAVING COUNT(*) > 1
|
List the name of artworks that are not nominated.
|
CREATE TABLE nomination (Name VARCHAR, Artwork_ID VARCHAR); CREATE TABLE Artwork (Name VARCHAR, Artwork_ID VARCHAR)
|
SELECT Name FROM Artwork WHERE NOT Artwork_ID IN (SELECT Artwork_ID FROM nomination)
|
### Context: CREATE TABLE nomination (Name VARCHAR, Artwork_ID VARCHAR); CREATE TABLE Artwork (Name VARCHAR, Artwork_ID VARCHAR) ### Question: List the name of artworks that are not nominated. ### Answer: SELECT Name FROM Artwork WHERE NOT Artwork_ID IN (SELECT Artwork_ID FROM nomination)
|
Show the number of audience in year 2008 or 2010.
|
CREATE TABLE festival_detail (Num_of_Audience VARCHAR, YEAR VARCHAR)
|
SELECT Num_of_Audience FROM festival_detail WHERE YEAR = 2008 OR YEAR = 2010
|
### Context: CREATE TABLE festival_detail (Num_of_Audience VARCHAR, YEAR VARCHAR) ### Question: Show the number of audience in year 2008 or 2010. ### Answer: SELECT Num_of_Audience FROM festival_detail WHERE YEAR = 2008 OR YEAR = 2010
|
What are the total number of the audiences who visited any of the festivals?
|
CREATE TABLE festival_detail (Num_of_Audience INTEGER)
|
SELECT SUM(Num_of_Audience) FROM festival_detail
|
### Context: CREATE TABLE festival_detail (Num_of_Audience INTEGER) ### Question: What are the total number of the audiences who visited any of the festivals? ### Answer: SELECT SUM(Num_of_Audience) FROM festival_detail
|
In which year are there festivals both inside the 'United States' and outside the 'United States'?
|
CREATE TABLE festival_detail (YEAR VARCHAR, LOCATION VARCHAR)
|
SELECT YEAR FROM festival_detail WHERE LOCATION = 'United States' INTERSECT SELECT YEAR FROM festival_detail WHERE LOCATION <> 'United States'
|
### Context: CREATE TABLE festival_detail (YEAR VARCHAR, LOCATION VARCHAR) ### Question: In which year are there festivals both inside the 'United States' and outside the 'United States'? ### Answer: SELECT YEAR FROM festival_detail WHERE LOCATION = 'United States' INTERSECT SELECT YEAR FROM festival_detail WHERE LOCATION <> 'United States'
|
How many premises are there?
|
CREATE TABLE premises (Id VARCHAR)
|
SELECT COUNT(*) FROM premises
|
### Context: CREATE TABLE premises (Id VARCHAR) ### Question: How many premises are there? ### Answer: SELECT COUNT(*) FROM premises
|
What are all the distinct premise types?
|
CREATE TABLE premises (premises_type VARCHAR)
|
SELECT DISTINCT premises_type FROM premises
|
### Context: CREATE TABLE premises (premises_type VARCHAR) ### Question: What are all the distinct premise types? ### Answer: SELECT DISTINCT premises_type FROM premises
|
Find the types and details for all premises and order by the premise type.
|
CREATE TABLE premises (premises_type VARCHAR, premise_details VARCHAR)
|
SELECT premises_type, premise_details FROM premises ORDER BY premises_type
|
### Context: CREATE TABLE premises (premises_type VARCHAR, premise_details VARCHAR) ### Question: Find the types and details for all premises and order by the premise type. ### Answer: SELECT premises_type, premise_details FROM premises ORDER BY premises_type
|
Show each premise type and the number of premises in that type.
|
CREATE TABLE premises (premises_type VARCHAR)
|
SELECT premises_type, COUNT(*) FROM premises GROUP BY premises_type
|
### Context: CREATE TABLE premises (premises_type VARCHAR) ### Question: Show each premise type and the number of premises in that type. ### Answer: SELECT premises_type, COUNT(*) FROM premises GROUP BY premises_type
|
Show all distinct product categories along with the number of mailshots in each category.
|
CREATE TABLE mailshot_campaigns (product_category VARCHAR)
|
SELECT product_category, COUNT(*) FROM mailshot_campaigns GROUP BY product_category
|
### Context: CREATE TABLE mailshot_campaigns (product_category VARCHAR) ### Question: Show all distinct product categories along with the number of mailshots in each category. ### Answer: SELECT product_category, COUNT(*) FROM mailshot_campaigns GROUP BY product_category
|
Show the name and phone of the customer without any mailshot.
|
CREATE TABLE customers (customer_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR)
|
SELECT customer_name, customer_phone FROM customers WHERE NOT customer_id IN (SELECT customer_id FROM mailshot_customers)
|
### Context: CREATE TABLE customers (customer_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR) ### Question: Show the name and phone of the customer without any mailshot. ### Answer: SELECT customer_name, customer_phone FROM customers WHERE NOT customer_id IN (SELECT customer_id FROM mailshot_customers)
|
Show the name and phone for customers with a mailshot with outcome code 'No Response'.
|
CREATE TABLE customers (customer_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_id VARCHAR, outcome_code VARCHAR)
|
SELECT T1.customer_name, T1.customer_phone FROM customers AS T1 JOIN mailshot_customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.outcome_code = 'No Response'
|
### Context: CREATE TABLE customers (customer_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_id VARCHAR, outcome_code VARCHAR) ### Question: Show the name and phone for customers with a mailshot with outcome code 'No Response'. ### Answer: SELECT T1.customer_name, T1.customer_phone FROM customers AS T1 JOIN mailshot_customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.outcome_code = 'No Response'
|
Show the outcome code of mailshots along with the number of mailshots in each outcome code.
|
CREATE TABLE mailshot_customers (outcome_code VARCHAR)
|
SELECT outcome_code, COUNT(*) FROM mailshot_customers GROUP BY outcome_code
|
### Context: CREATE TABLE mailshot_customers (outcome_code VARCHAR) ### Question: Show the outcome code of mailshots along with the number of mailshots in each outcome code. ### Answer: SELECT outcome_code, COUNT(*) FROM mailshot_customers GROUP BY outcome_code
|
Show the names of customers who have at least 2 mailshots with outcome code 'Order'.
|
CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_id VARCHAR)
|
SELECT T2.customer_name FROM mailshot_customers AS T1 JOIN customers AS T2 ON T1.customer_id = T2.customer_id WHERE outcome_code = 'Order' GROUP BY T1.customer_id HAVING COUNT(*) >= 2
|
### Context: CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_id VARCHAR) ### Question: Show the names of customers who have at least 2 mailshots with outcome code 'Order'. ### Answer: SELECT T2.customer_name FROM mailshot_customers AS T1 JOIN customers AS T2 ON T1.customer_id = T2.customer_id WHERE outcome_code = 'Order' GROUP BY T1.customer_id HAVING COUNT(*) >= 2
|
Show the names of customers who have the most mailshots.
|
CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_id VARCHAR)
|
SELECT T2.customer_name FROM mailshot_customers AS T1 JOIN customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) DESC LIMIT 1
|
### Context: CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_id VARCHAR) ### Question: Show the names of customers who have the most mailshots. ### Answer: SELECT T2.customer_name FROM mailshot_customers AS T1 JOIN customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) DESC LIMIT 1
|
What are the name and payment method of customers who have both mailshots in 'Order' outcome and mailshots in 'No Response' outcome.
|
CREATE TABLE customers (customer_name VARCHAR, payment_method VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_id VARCHAR, outcome_code VARCHAR)
|
SELECT T2.customer_name, T2.payment_method FROM mailshot_customers AS T1 JOIN customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.outcome_code = 'Order' INTERSECT SELECT T2.customer_name, T2.payment_method FROM mailshot_customers AS T1 JOIN customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.outcome_code = 'No Response'
|
### Context: CREATE TABLE customers (customer_name VARCHAR, payment_method VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_id VARCHAR, outcome_code VARCHAR) ### Question: What are the name and payment method of customers who have both mailshots in 'Order' outcome and mailshots in 'No Response' outcome. ### Answer: SELECT T2.customer_name, T2.payment_method FROM mailshot_customers AS T1 JOIN customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.outcome_code = 'Order' INTERSECT SELECT T2.customer_name, T2.payment_method FROM mailshot_customers AS T1 JOIN customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.outcome_code = 'No Response'
|
Show the premise type and address type code for all customer addresses.
|
CREATE TABLE premises (premises_type VARCHAR, premise_id VARCHAR); CREATE TABLE customer_addresses (address_type_code VARCHAR, premise_id VARCHAR)
|
SELECT T2.premises_type, T1.address_type_code FROM customer_addresses AS T1 JOIN premises AS T2 ON T1.premise_id = T2.premise_id
|
### Context: CREATE TABLE premises (premises_type VARCHAR, premise_id VARCHAR); CREATE TABLE customer_addresses (address_type_code VARCHAR, premise_id VARCHAR) ### Question: Show the premise type and address type code for all customer addresses. ### Answer: SELECT T2.premises_type, T1.address_type_code FROM customer_addresses AS T1 JOIN premises AS T2 ON T1.premise_id = T2.premise_id
|
What are the distinct address type codes for all customer addresses?
|
CREATE TABLE customer_addresses (address_type_code VARCHAR)
|
SELECT DISTINCT address_type_code FROM customer_addresses
|
### Context: CREATE TABLE customer_addresses (address_type_code VARCHAR) ### Question: What are the distinct address type codes for all customer addresses? ### Answer: SELECT DISTINCT address_type_code FROM customer_addresses
|
Show the shipping charge and customer id for customer orders with order status Cancelled or Paid.
|
CREATE TABLE customer_orders (order_shipping_charges VARCHAR, customer_id VARCHAR, order_status_code VARCHAR)
|
SELECT order_shipping_charges, customer_id FROM customer_orders WHERE order_status_code = 'Cancelled' OR order_status_code = 'Paid'
|
### Context: CREATE TABLE customer_orders (order_shipping_charges VARCHAR, customer_id VARCHAR, order_status_code VARCHAR) ### Question: Show the shipping charge and customer id for customer orders with order status Cancelled or Paid. ### Answer: SELECT order_shipping_charges, customer_id FROM customer_orders WHERE order_status_code = 'Cancelled' OR order_status_code = 'Paid'
|
Show the names of customers having an order with shipping method FedEx and order status Paid.
|
CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE customer_orders (customer_id VARCHAR)
|
SELECT T1.customer_name FROM customers AS T1 JOIN customer_orders AS T2 ON T1.customer_id = T2.customer_id WHERE shipping_method_code = 'FedEx' AND order_status_code = 'Paid'
|
### Context: CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE customer_orders (customer_id VARCHAR) ### Question: Show the names of customers having an order with shipping method FedEx and order status Paid. ### Answer: SELECT T1.customer_name FROM customers AS T1 JOIN customer_orders AS T2 ON T1.customer_id = T2.customer_id WHERE shipping_method_code = 'FedEx' AND order_status_code = 'Paid'
|
How many courses are there in total?
|
CREATE TABLE COURSE (Id VARCHAR)
|
SELECT COUNT(*) FROM COURSE
|
### Context: CREATE TABLE COURSE (Id VARCHAR) ### Question: How many courses are there in total? ### Answer: SELECT COUNT(*) FROM COURSE
|
How many courses have more than 2 credits?
|
CREATE TABLE COURSE (Credits INTEGER)
|
SELECT COUNT(*) FROM COURSE WHERE Credits > 2
|
### Context: CREATE TABLE COURSE (Credits INTEGER) ### Question: How many courses have more than 2 credits? ### Answer: SELECT COUNT(*) FROM COURSE WHERE Credits > 2
|
List all names of courses with 1 credit?
|
CREATE TABLE COURSE (CName VARCHAR, Credits VARCHAR)
|
SELECT CName FROM COURSE WHERE Credits = 1
|
### Context: CREATE TABLE COURSE (CName VARCHAR, Credits VARCHAR) ### Question: List all names of courses with 1 credit? ### Answer: SELECT CName FROM COURSE WHERE Credits = 1
|
Which courses are taught on days MTW?
|
CREATE TABLE COURSE (CName VARCHAR, Days VARCHAR)
|
SELECT CName FROM COURSE WHERE Days = "MTW"
|
### Context: CREATE TABLE COURSE (CName VARCHAR, Days VARCHAR) ### Question: Which courses are taught on days MTW? ### Answer: SELECT CName FROM COURSE WHERE Days = "MTW"
|
What is the number of departments in Division "AS"?
|
CREATE TABLE DEPARTMENT (Division VARCHAR)
|
SELECT COUNT(*) FROM DEPARTMENT WHERE Division = "AS"
|
### Context: CREATE TABLE DEPARTMENT (Division VARCHAR) ### Question: What is the number of departments in Division "AS"? ### Answer: SELECT COUNT(*) FROM DEPARTMENT WHERE Division = "AS"
|
What are the phones of departments in Room 268?
|
CREATE TABLE DEPARTMENT (DPhone VARCHAR, Room VARCHAR)
|
SELECT DPhone FROM DEPARTMENT WHERE Room = 268
|
### Context: CREATE TABLE DEPARTMENT (DPhone VARCHAR, Room VARCHAR) ### Question: What are the phones of departments in Room 268? ### Answer: SELECT DPhone FROM DEPARTMENT WHERE Room = 268
|
Find the number of students that have at least one grade "B".
|
CREATE TABLE ENROLLED_IN (StuID VARCHAR, Grade VARCHAR)
|
SELECT COUNT(DISTINCT StuID) FROM ENROLLED_IN WHERE Grade = "B"
|
### Context: CREATE TABLE ENROLLED_IN (StuID VARCHAR, Grade VARCHAR) ### Question: Find the number of students that have at least one grade "B". ### Answer: SELECT COUNT(DISTINCT StuID) FROM ENROLLED_IN WHERE Grade = "B"
|
Find the max and min grade point for all letter grade.
|
CREATE TABLE GRADECONVERSION (gradepoint INTEGER)
|
SELECT MAX(gradepoint), MIN(gradepoint) FROM GRADECONVERSION
|
### Context: CREATE TABLE GRADECONVERSION (gradepoint INTEGER) ### Question: Find the max and min grade point for all letter grade. ### Answer: SELECT MAX(gradepoint), MIN(gradepoint) FROM GRADECONVERSION
|
Find the first names of students whose first names contain letter "a".
|
CREATE TABLE STUDENT (Fname VARCHAR)
|
SELECT DISTINCT Fname FROM STUDENT WHERE Fname LIKE '%a%'
|
### Context: CREATE TABLE STUDENT (Fname VARCHAR) ### Question: Find the first names of students whose first names contain letter "a". ### Answer: SELECT DISTINCT Fname FROM STUDENT WHERE Fname LIKE '%a%'
|
Find the first names and last names of male (sex is M) faculties who live in building NEB.
|
CREATE TABLE FACULTY (Fname VARCHAR, Lname VARCHAR, sex VARCHAR, Building VARCHAR)
|
SELECT Fname, Lname FROM FACULTY WHERE sex = "M" AND Building = "NEB"
|
### Context: CREATE TABLE FACULTY (Fname VARCHAR, Lname VARCHAR, sex VARCHAR, Building VARCHAR) ### Question: Find the first names and last names of male (sex is M) faculties who live in building NEB. ### Answer: SELECT Fname, Lname FROM FACULTY WHERE sex = "M" AND Building = "NEB"
|
Find the rooms of faculties with rank professor who live in building NEB.
|
CREATE TABLE FACULTY (Room VARCHAR, Rank VARCHAR, Building VARCHAR)
|
SELECT Room FROM FACULTY WHERE Rank = "Professor" AND Building = "NEB"
|
### Context: CREATE TABLE FACULTY (Room VARCHAR, Rank VARCHAR, Building VARCHAR) ### Question: Find the rooms of faculties with rank professor who live in building NEB. ### Answer: SELECT Room FROM FACULTY WHERE Rank = "Professor" AND Building = "NEB"
|
Find the department name that is in Building "Mergenthaler".
|
CREATE TABLE DEPARTMENT (DName VARCHAR, Building VARCHAR)
|
SELECT DName FROM DEPARTMENT WHERE Building = "Mergenthaler"
|
### Context: CREATE TABLE DEPARTMENT (DName VARCHAR, Building VARCHAR) ### Question: Find the department name that is in Building "Mergenthaler". ### Answer: SELECT DName FROM DEPARTMENT WHERE Building = "Mergenthaler"
|
List all information about courses sorted by credits in the ascending order.
|
CREATE TABLE COURSE (Credits VARCHAR)
|
SELECT * FROM COURSE ORDER BY Credits
|
### Context: CREATE TABLE COURSE (Credits VARCHAR) ### Question: List all information about courses sorted by credits in the ascending order. ### Answer: SELECT * FROM COURSE ORDER BY Credits
|
List the course name of courses sorted by credits.
|
CREATE TABLE COURSE (CName VARCHAR, Credits VARCHAR)
|
SELECT CName FROM COURSE ORDER BY Credits
|
### Context: CREATE TABLE COURSE (CName VARCHAR, Credits VARCHAR) ### Question: List the course name of courses sorted by credits. ### Answer: SELECT CName FROM COURSE ORDER BY Credits
|
Find the first name of students in the descending order of age.
|
CREATE TABLE STUDENT (Fname VARCHAR, Age VARCHAR)
|
SELECT Fname FROM STUDENT ORDER BY Age DESC
|
### Context: CREATE TABLE STUDENT (Fname VARCHAR, Age VARCHAR) ### Question: Find the first name of students in the descending order of age. ### Answer: SELECT Fname FROM STUDENT ORDER BY Age DESC
|
Find the last name of female (sex is F) students in the descending order of age.
|
CREATE TABLE STUDENT (LName VARCHAR, Sex VARCHAR, Age VARCHAR)
|
SELECT LName FROM STUDENT WHERE Sex = "F" ORDER BY Age DESC
|
### Context: CREATE TABLE STUDENT (LName VARCHAR, Sex VARCHAR, Age VARCHAR) ### Question: Find the last name of female (sex is F) students in the descending order of age. ### Answer: SELECT LName FROM STUDENT WHERE Sex = "F" ORDER BY Age DESC
|
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