input
stringlengths 236
16.9k
| output
stringlengths 19
805
|
|---|---|
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How many females does this network has?
| SELECT count(*) FROM Person WHERE gender = 'female'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How many females are in the network?
| SELECT count(*) FROM Person WHERE gender = 'female'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the average age for all person?
| SELECT avg(age) FROM Person; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the average age for all people in the table?
| SELECT avg(age) FROM Person; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How many different cities are they from?
| SELECT count(DISTINCT city) FROM Person; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How many different cities do people originate from?
| SELECT count(DISTINCT city) FROM Person; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How many type of jobs do they have?
| SELECT count(DISTINCT job) FROM Person; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How many different jobs are listed?
| SELECT count(DISTINCT job) FROM Person; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Who is the oldest person?
| SELECT name FROM Person WHERE age = (SELECT max(age) FROM person); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the name of the person who is the oldest?
| SELECT name FROM Person WHERE age = (SELECT max(age) FROM person); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Who is the oldest person whose job is student?
| SELECT name FROM Person WHERE job = 'student' AND age = (SELECT max(age) FROM person WHERE job = 'student' ); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the name of the oldest student?
| SELECT name FROM Person WHERE job = 'student' AND age = (SELECT max(age) FROM person WHERE job = 'student' ); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Who is the youngest male?
| SELECT name FROM Person WHERE gender = 'male' AND age = (SELECT min(age) FROM person WHERE gender = 'male' ); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the name of the youngest male?
| SELECT name FROM Person WHERE gender = 'male' AND age = (SELECT min(age) FROM person WHERE gender = 'male' ); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How old is the doctor named Zach?
| SELECT age FROM Person WHERE job = 'doctor' AND name = 'Zach'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the age of the doctor named Zach?
| SELECT age FROM Person WHERE job = 'doctor' AND name = 'Zach'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Who is the person whose age is below 30?
| SELECT name FROM Person WHERE age < 30; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the name of the person whose age is below 30?
| SELECT name FROM Person WHERE age < 30; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How many people whose age is greater 30 and job is engineer?
| SELECT count(*) FROM Person WHERE age > 30 AND job = 'engineer'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- HOw many engineers are older than 30?
| SELECT count(*) FROM Person WHERE age > 30 AND job = 'engineer'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the average age for each gender?
| SELECT avg(age) , gender FROM Person GROUP BY gender; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How old is each gender, on average?
| SELECT avg(age) , gender FROM Person GROUP BY gender; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is average age for different job title?
| SELECT avg(age) , job FROM Person GROUP BY job; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How old is the average person for each job?
| SELECT avg(age) , job FROM Person GROUP BY job; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is average age of male for different job title?
| SELECT avg(age) , job FROM Person WHERE gender = 'male' GROUP BY job; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the average age for a male in each job?
| SELECT avg(age) , job FROM Person WHERE gender = 'male' GROUP BY job; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is minimum age for different job title?
| SELECT min(age) , job FROM Person GROUP BY job; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How old is the youngest person for each job?
| SELECT min(age) , job FROM Person GROUP BY job; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the number of people who is under 40 for each gender.
| SELECT count(*) , gender FROM Person WHERE age < 40 GROUP BY gender; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How many people are under 40 for each gender?
| SELECT count(*) , gender FROM Person WHERE age < 40 GROUP BY gender; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the name of people whose age is greater than any engineer sorted by their age.
| SELECT name FROM Person WHERE age > (SELECT min(age) FROM person WHERE job = 'engineer') ORDER BY age; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the name of all the people who are older than at least one engineer? Order them by age.
| SELECT name FROM Person WHERE age > (SELECT min(age) FROM person WHERE job = 'engineer') ORDER BY age; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the number of people whose age is greater than all engineers.
| SELECT count(*) FROM Person WHERE age > (SELECT max(age) FROM person WHERE job = 'engineer'); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- How many people are older than every engineer?
| SELECT count(*) FROM Person WHERE age > (SELECT max(age) FROM person WHERE job = 'engineer'); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- list the name, job title of all people ordered by their names.
| SELECT name , job FROM Person ORDER BY name; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names and job titles of every person ordered alphabetically by name?
| SELECT name , job FROM Person ORDER BY name; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the names of all person sorted in the descending order using age.
| SELECT name FROM Person ORDER BY age DESC; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names of everybody sorted by age in descending order?
| SELECT name FROM Person ORDER BY age DESC; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the name and age of all males in order of their age.
| SELECT name FROM Person WHERE gender = 'male' ORDER BY age; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the name and age of every male? Order the results by age.
| SELECT name FROM Person WHERE gender = 'male' ORDER BY age; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the name and age of the person who is a friend of both Dan and Alice.
| SELECT T1.name , T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Dan' INTERSECT SELECT T1.name , T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Alice'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names and ages of every person who is a friend of both Dan and Alice?
| SELECT T1.name , T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Dan' INTERSECT SELECT T1.name , T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Alice'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the name and age of the person who is a friend of Dan or Alice.
| SELECT DISTINCT T1.name , T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Dan' OR T2.friend = 'Alice'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the different names and ages of every friend of either Dan or alice?
| SELECT DISTINCT T1.name , T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Dan' OR T2.friend = 'Alice'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the name of the person who has friends with age above 40 and under age 30?
| SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend IN (SELECT name FROM Person WHERE age > 40) INTERSECT SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend IN (SELECT name FROM Person WHERE age < 30); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names of every person who has a friend over 40 and under 30?
| SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend IN (SELECT name FROM Person WHERE age > 40) INTERSECT SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend IN (SELECT name FROM Person WHERE age < 30); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the name of the person who has friends with age above 40 but not under age 30?
| SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend IN (SELECT name FROM Person WHERE age > 40) EXCEPT SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend IN (SELECT name FROM Person WHERE age < 30); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names of the people who are older 40 but no friends under age 30?
| SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend IN (SELECT name FROM Person WHERE age > 40) EXCEPT SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend IN (SELECT name FROM Person WHERE age < 30); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the name of the person who has no student friends.
| SELECT name FROM person EXCEPT SELECT T2.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.job = 'student'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names of the people who have no friends who are students?
| SELECT name FROM person EXCEPT SELECT T2.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.job = 'student'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the person who has exactly one friend.
| SELECT name FROM PersonFriend GROUP BY name HAVING count(*) = 1; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names of everybody who has exactly one friend?
| SELECT name FROM PersonFriend GROUP BY name HAVING count(*) = 1; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Who are the friends of Bob?
| SELECT T2.friend FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T1.name = 'Bob'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Who are Bob's friends?
| SELECT T2.friend FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T1.name = 'Bob'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the name of persons who are friends with Bob.
| SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Bob'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names of all of Bob's friends?
| SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Bob'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the names of females who are friends with Zach
| SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Zach' AND T1.gender = 'female'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names of all females who are friends with Zach?
| SELECT T1.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Zach' AND T1.gender = 'female'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the female friends of Alice.
| SELECT T2.friend FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T2.name = 'Alice' AND T1.gender = 'female'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are all the friends of Alice who are female?
| SELECT T2.friend FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T2.name = 'Alice' AND T1.gender = 'female'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the male friend of Alice whose job is a doctor?
| SELECT T2.friend FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T2.name = 'Alice' AND T1.gender = 'male' AND T1.job = 'doctor'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Who are the friends of Alice that are doctors?
| SELECT T2.friend FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T2.name = 'Alice' AND T1.gender = 'male' AND T1.job = 'doctor'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Who has a friend that is from new york city?
| SELECT T2.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.city = 'new york city'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names of all friends who are from New York?
| SELECT T2.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.city = 'new york city'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Who has friends that are younger than the average age?
| SELECT DISTINCT T2.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.age < (SELECT avg(age) FROM person); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the different names of friends who are younger than the average age for a friend?
| SELECT DISTINCT T2.name FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.age < (SELECT avg(age) FROM person); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Who has friends that are older than the average age? Print their friends and their ages as well
| SELECT DISTINCT T2.name , T2.friend , T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.age > (SELECT avg(age) FROM person); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Whare the names, friends, and ages of all people who are older than the average age of a person?
| SELECT DISTINCT T2.name , T2.friend , T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T1.age > (SELECT avg(age) FROM person); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Who is the friend of Zach with longest year relationship?
| SELECT friend FROM PersonFriend WHERE name = 'Zach' AND YEAR = (SELECT max(YEAR) FROM PersonFriend WHERE name = 'Zach'); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Which friend of Zach has the longest-lasting friendship?
| SELECT friend FROM PersonFriend WHERE name = 'Zach' AND YEAR = (SELECT max(YEAR) FROM PersonFriend WHERE name = 'Zach'); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the age of the friend of Zach with longest year relationship?
| SELECT T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T2.name = 'Zach' AND T2.year = (SELECT max(YEAR) FROM PersonFriend WHERE name = 'Zach'); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the ages of all of Zach's friends who are in the longest relationship?
| SELECT T1.age FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend WHERE T2.name = 'Zach' AND T2.year = (SELECT max(YEAR) FROM PersonFriend WHERE name = 'Zach'); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the name of persons who are friends with Alice for the shortest years.
| SELECT name FROM PersonFriend WHERE friend = 'Alice' AND YEAR = (SELECT min(YEAR) FROM PersonFriend WHERE friend = 'Alice'); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names of all people who are friends with Alice for the shortest amount of time?
| SELECT name FROM PersonFriend WHERE friend = 'Alice' AND YEAR = (SELECT min(YEAR) FROM PersonFriend WHERE friend = 'Alice'); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find the name, age, and job title of persons who are friends with Alice for the longest years.
| SELECT T1.name , T1.age , T1.job FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Alice' AND T2.year = (SELECT max(YEAR) FROM PersonFriend WHERE friend = 'Alice'); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names, ages, and jobs of all people who are friends with Alice for the longest amount of time?
| SELECT T1.name , T1.age , T1.job FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T2.friend = 'Alice' AND T2.year = (SELECT max(YEAR) FROM PersonFriend WHERE friend = 'Alice'); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Who is the person that has no friend?
| SELECT name FROM person EXCEPT SELECT name FROM PersonFriend; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names of all people who do not have friends?
| SELECT name FROM person EXCEPT SELECT name FROM PersonFriend; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Which person whose friends have the oldest average age?
| SELECT T2.name , avg(T1.age) FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend GROUP BY T2.name ORDER BY avg(T1.age) DESC LIMIT 1; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the name of the person who has the oldest average age for their friends, and what is that average age?
| SELECT T2.name , avg(T1.age) FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.friend GROUP BY T2.name ORDER BY avg(T1.age) DESC LIMIT 1; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the total number of people who has no friend living in the city of Austin.
| SELECT count(DISTINCT name) FROM PersonFriend WHERE friend NOT IN (SELECT name FROM person WHERE city = 'Austin'); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What is the total number of people who have no friends living in Austin?
| SELECT count(DISTINCT name) FROM PersonFriend WHERE friend NOT IN (SELECT name FROM person WHERE city = 'Austin'); |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- Find Alice's friends of friends.
| SELECT DISTINCT T4.name FROM PersonFriend AS T1 JOIN Person AS T2 ON T1.name = T2.name JOIN PersonFriend AS T3 ON T1.friend = T3.name JOIN PersonFriend AS T4 ON T3.friend = T4.name WHERE T2.name = 'Alice' AND T4.name != 'Alice'; |
-- Database schema
| Person : name [ TEXT ] primary_key , age [ INT ] , city [ TEXT ] , gender [ TEXT ] , job [ TEXT ] | PersonFriend : name [ TEXT ] PersonFriend.name = Person.name , friend [ TEXT ] PersonFriend.friend = Person.name , year [ INT ] |
-- -- What are the names of all of Alice's friends of friends?
| SELECT DISTINCT T4.name FROM PersonFriend AS T1 JOIN Person AS T2 ON T1.name = T2.name JOIN PersonFriend AS T3 ON T1.friend = T3.name JOIN PersonFriend AS T4 ON T3.friend = T4.name WHERE T2.name = 'Alice' AND T4.name != 'Alice'; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- How many members are there?
| SELECT count(*) FROM member; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- List the names of members in ascending alphabetical order.
| SELECT Name FROM member ORDER BY Name ASC; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- What are the names and countries of members?
| SELECT Name , Country FROM member; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- Show the names of members whose country is "United States" or "Canada".
| SELECT Name FROM member WHERE Country = "United States" OR Country = "Canada"; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- Show the different countries and the number of members from each.
| SELECT Country , COUNT(*) FROM member GROUP BY Country; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- Show the most common country across members.
| SELECT Country FROM member GROUP BY Country ORDER BY COUNT(*) DESC LIMIT 1; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- Which countries have more than two members?
| SELECT Country FROM member GROUP BY Country HAVING COUNT(*) > 2; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- Show the leader names and locations of colleges.
| SELECT Leader_Name , College_Location FROM college; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- Show the names of members and names of colleges they go to.
| SELECT T2.Name , T1.Name FROM college AS T1 JOIN member AS T2 ON T1.College_ID = T2.College_ID; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- Show the names of members and the locations of colleges they go to in ascending alphabetical order of member names.
| SELECT T2.Name , T1.College_Location FROM college AS T1 JOIN member AS T2 ON T1.College_ID = T2.College_ID ORDER BY T2.Name ASC; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- Show the distinct leader names of colleges associated with members from country "Canada".
| SELECT DISTINCT T1.Leader_Name FROM college AS T1 JOIN member AS T2 ON T1.College_ID = T2.College_ID WHERE T2.Country = "Canada"; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- Show the names of members and the decoration themes they have.
| SELECT T1.Name , T2.Decoration_Theme FROM member AS T1 JOIN round AS T2 ON T1.Member_ID = T2.Member_ID; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- Show the names of members that have a rank in round higher than 3.
| SELECT T1.Name FROM member AS T1 JOIN round AS T2 ON T1.Member_ID = T2.Member_ID WHERE T2.Rank_in_Round > 3; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- Show the names of members in ascending order of their rank in rounds.
| SELECT T1.Name FROM member AS T1 JOIN round AS T2 ON T1.Member_ID = T2.Member_ID ORDER BY Rank_in_Round ASC; |
-- Database schema
| college : College_ID [ INT ] primary_key , Name [ TEXT ] , Leader_Name [ TEXT ] , College_Location [ TEXT ] | member : Member_ID [ INT ] primary_key , Name [ TEXT ] , Country [ TEXT ] , College_ID [ INT ] member.College_ID = college.College_ID | round : Round_ID [ INT ] , Member_ID [ INT ] primary_key round.Member_ID = member.Member_ID , Decoration_Theme [ TEXT ] , Rank_in_Round [ INT ] |
-- -- List the names of members who did not participate in any round.
| SELECT Name FROM member WHERE Member_ID NOT IN (SELECT Member_ID FROM round); |
-- Database schema
| Roles : role_code [ TEXT ] primary_key , role_description [ TEXT ] | Users : user_id [ INT ] primary_key , role_code [ TEXT ] Users.role_code = Roles.role_code , user_name [ TEXT ] , user_login [ TEXT ] , password [ TEXT ] | Document_Structures : document_structure_code [ TEXT ] primary_key , parent_document_structure_code [ TEXT ] , document_structure_description [ TEXT ] | Functional_Areas : functional_area_code [ TEXT ] primary_key , parent_functional_area_code [ TEXT ] , functional_area_description [ TEXT ] | Images : image_id [ INT ] primary_key , image_alt_text [ TEXT ] , image_name [ TEXT ] , image_url [ TEXT ] | Documents : document_code [ TEXT ] primary_key , document_structure_code [ TEXT ] Documents.document_structure_code = Document_Structures.document_structure_code , document_type_code [ TEXT ] , access_count [ INT ] , document_name [ TEXT ] | Document_Functional_Areas : document_code [ TEXT ] Document_Functional_Areas.document_code = Documents.document_code , functional_area_code [ TEXT ] Document_Functional_Areas.functional_area_code = Functional_Areas.functional_area_code | Document_Sections : section_id [ INT ] primary_key , document_code [ TEXT ] Document_Sections.document_code = Documents.document_code , section_sequence [ INT ] , section_code [ TEXT ] , section_title [ TEXT ] | Document_Sections_Images : section_id [ INT ] primary_key Document_Sections_Images.section_id = Document_Sections.section_id , image_id [ INT ] Document_Sections_Images.image_id = Images.image_id |
-- -- Find the name and access counts of all documents, in alphabetic order of the document name.
| SELECT document_name , access_count FROM documents ORDER BY document_name; |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.