Datasets:
AayushShah
/
Univeral_SQL_Three_Datasets_Combined_WithText_IDs

Dataset card Data Studio Files
xet
Community
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244
SQL
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List the project details of the projects launched by the organisation
SELECT project_details FROM Projects WHERE organisation_id IN (SELECT organisation_id FROM Projects GROUP BY organisation_id ORDER BY COUNT(*) DESC LIMIT 1)
CREATE TABLE Projects (project_details VARCHAR, organisation_id VARCHAR)
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List the research staff details, and order in ascending order.
SELECT staff_details FROM Research_Staff ORDER BY staff_details
CREATE TABLE Research_Staff (staff_details VARCHAR)
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How many tasks are there in total?
SELECT COUNT(*) FROM Tasks
CREATE TABLE Tasks (Id VARCHAR)
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How many tasks does each project have? List the task count and the project detail.
SELECT COUNT(*), T1.project_details FROM Projects AS T1 JOIN Tasks AS T2 ON T1.project_id = T2.project_id GROUP BY T1.project_id
CREATE TABLE Projects (project_details VARCHAR, project_id VARCHAR); CREATE TABLE Tasks (project_id VARCHAR)
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What are the staff roles of the staff who
SELECT role_code FROM Project_Staff WHERE date_from > '2003-04-19 15:06:20' AND date_to < '2016-03-15 00:33:18'
CREATE TABLE Project_Staff (role_code VARCHAR, date_from VARCHAR, date_to VARCHAR)
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What are the descriptions of all the project outcomes?
SELECT T1.outcome_description FROM Research_outcomes AS T1 JOIN Project_outcomes AS T2 ON T1.outcome_code = T2.outcome_code
CREATE TABLE Research_outcomes (outcome_description VARCHAR, outcome_code VARCHAR); CREATE TABLE Project_outcomes (outcome_code VARCHAR)
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Which role is most common for the staff?
SELECT role_code FROM Project_Staff GROUP BY role_code ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Project_Staff (role_code VARCHAR)
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How many friends does Dan have?
SELECT COUNT(T2.friend) FROM Person AS T1 JOIN PersonFriend AS T2 ON T1.name = T2.name WHERE T1.name = 'Dan'
CREATE TABLE PersonFriend (friend VARCHAR, name VARCHAR); CREATE TABLE Person (name VARCHAR)
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How many females does this network has?
SELECT COUNT(*) FROM Person WHERE gender = 'female'
CREATE TABLE Person (gender VARCHAR)
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What is the average age for all person?
SELECT AVG(age) FROM Person
CREATE TABLE Person (age INTEGER)
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How many different cities are they from?
SELECT COUNT(DISTINCT city) FROM Person
CREATE TABLE Person (city VARCHAR)
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How many type of jobs do they have?
SELECT COUNT(DISTINCT job) FROM Person
CREATE TABLE Person (job VARCHAR)
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Who is the oldest person?
SELECT name FROM Person WHERE age = (SELECT MAX(age) FROM person)
CREATE TABLE Person (name VARCHAR, age INTEGER); CREATE TABLE person (name VARCHAR, age INTEGER)
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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')
CREATE TABLE person (name VARCHAR, job VARCHAR, age INTEGER); CREATE TABLE Person (name VARCHAR, job VARCHAR, age INTEGER)
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Who is the youngest male?
SELECT name FROM Person WHERE gender = 'male' AND age = (SELECT MIN(age) FROM person WHERE gender = 'male')
CREATE TABLE Person (name VARCHAR, gender VARCHAR, age INTEGER); CREATE TABLE person (name VARCHAR, gender VARCHAR, age INTEGER)
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How old is the doctor named Zach?
SELECT age FROM Person WHERE job = 'doctor' AND name = 'Zach'
CREATE TABLE Person (age VARCHAR, job VARCHAR, name VARCHAR)
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Who is the person whose age is below 30?
SELECT name FROM Person WHERE age < 30
CREATE TABLE Person (name VARCHAR, age INTEGER)
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How many people whose age is greater 30 and job is engineer?
SELECT COUNT(*) FROM Person WHERE age > 30 AND job = 'engineer'
CREATE TABLE Person (age VARCHAR, job VARCHAR)
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What is the average age for each gender?
SELECT AVG(age), gender FROM Person GROUP BY gender
CREATE TABLE Person (gender VARCHAR, age INTEGER)
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What is average age for different job title?
SELECT AVG(age), job FROM Person GROUP BY job
CREATE TABLE Person (job VARCHAR, age INTEGER)
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What is average age of male for different job title?
SELECT AVG(age), job FROM Person WHERE gender = 'male' GROUP BY job
CREATE TABLE Person (job VARCHAR, age INTEGER, gender VARCHAR)
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What is minimum age for different job title?
SELECT MIN(age), job FROM Person GROUP BY job
CREATE TABLE Person (job VARCHAR, age INTEGER)
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Find the number of people who is under 40 for each gender.
SELECT COUNT(*), gender FROM Person WHERE age < 40 GROUP BY gender
CREATE TABLE Person (gender VARCHAR, age INTEGER)
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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
CREATE TABLE Person (name VARCHAR, age INTEGER, job VARCHAR); CREATE TABLE person (name VARCHAR, age INTEGER, job VARCHAR)
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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')
CREATE TABLE Person (age INTEGER, job VARCHAR); CREATE TABLE person (age INTEGER, job VARCHAR)
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list the name, job title of all people ordered by their names.
SELECT name, job FROM Person ORDER BY name
CREATE TABLE Person (name VARCHAR, job VARCHAR)
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Find the names of all person sorted in the descending order using age.
SELECT name FROM Person ORDER BY age DESC
CREATE TABLE Person (name VARCHAR, age VARCHAR)
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Find the name and age of all males in order of their age.
SELECT name FROM Person WHERE gender = 'male' ORDER BY age
CREATE TABLE Person (name VARCHAR, gender VARCHAR, age VARCHAR)
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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'
CREATE TABLE Person (name VARCHAR, age VARCHAR); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR)
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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'
CREATE TABLE Person (name VARCHAR, age VARCHAR); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR)
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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)
CREATE TABLE Person (name VARCHAR, age INTEGER); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (name VARCHAR)
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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)
CREATE TABLE Person (name VARCHAR, age INTEGER); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (name VARCHAR)
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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'
CREATE TABLE Person (name VARCHAR, job VARCHAR); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE person (name VARCHAR)
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Find the person who has exactly one friend.
SELECT name FROM PersonFriend GROUP BY name HAVING COUNT(*) = 1
CREATE TABLE PersonFriend (name VARCHAR)
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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'
CREATE TABLE PersonFriend (friend VARCHAR, name VARCHAR); CREATE TABLE Person (name VARCHAR)
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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'
CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (name VARCHAR)
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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'
CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (name VARCHAR, gender VARCHAR)
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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'
CREATE TABLE PersonFriend (friend VARCHAR, name VARCHAR); CREATE TABLE Person (name VARCHAR, gender VARCHAR)
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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'
CREATE TABLE PersonFriend (friend VARCHAR, name VARCHAR); CREATE TABLE Person (name VARCHAR, job VARCHAR, gender VARCHAR)
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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'
CREATE TABLE Person (name VARCHAR, city VARCHAR); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR)
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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)
CREATE TABLE person (age INTEGER); CREATE TABLE Person (name VARCHAR, age INTEGER); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR)
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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)
CREATE TABLE person (age INTEGER); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (age INTEGER, name VARCHAR)
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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')
CREATE TABLE PersonFriend (friend VARCHAR, name VARCHAR, YEAR INTEGER)
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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')
CREATE TABLE PersonFriend (friend VARCHAR, name VARCHAR, year VARCHAR); CREATE TABLE PersonFriend (YEAR INTEGER, name VARCHAR); CREATE TABLE Person (age VARCHAR, name VARCHAR)
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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')
CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR, YEAR INTEGER)
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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')
CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR, year VARCHAR); CREATE TABLE Person (name VARCHAR, age VARCHAR, job VARCHAR); CREATE TABLE PersonFriend (YEAR INTEGER, friend VARCHAR)
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Who is the person that has no friend?
SELECT name FROM person EXCEPT SELECT name FROM PersonFriend
CREATE TABLE PersonFriend (name VARCHAR); CREATE TABLE person (name VARCHAR)
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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
CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (age INTEGER, name VARCHAR)
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What is the total number of people who has no friend living in the city of Austin.
SELECT COUNT(DISTINCT name) FROM PersonFriend WHERE NOT friend IN (SELECT name FROM person WHERE city = 'Austin')
CREATE TABLE person (name VARCHAR, friend VARCHAR, city VARCHAR); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR, city VARCHAR)
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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'
CREATE TABLE PersonFriend (name VARCHAR); CREATE TABLE PersonFriend (name VARCHAR, friend VARCHAR); CREATE TABLE Person (name VARCHAR)
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How many members are there?
SELECT COUNT(*) FROM member
CREATE TABLE member (Id VARCHAR)
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List the names of members in ascending alphabetical order.
SELECT Name FROM member ORDER BY Name
CREATE TABLE member (Name VARCHAR)
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What are the names and countries of members?
SELECT Name, Country FROM member
CREATE TABLE member (Name VARCHAR, Country VARCHAR)
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Show the names of members whose country is "United States" or "Canada".
SELECT Name FROM member WHERE Country = "United States" OR Country = "Canada"
CREATE TABLE member (Name VARCHAR, Country VARCHAR)
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Show the different countries and the number of members from each.
SELECT Country, COUNT(*) FROM member GROUP BY Country
CREATE TABLE member (Country VARCHAR)
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Show the most common country across members.
SELECT Country FROM member GROUP BY Country ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE member (Country VARCHAR)
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Which countries have more than two members?
SELECT Country FROM member GROUP BY Country HAVING COUNT(*) > 2
CREATE TABLE member (Country VARCHAR)
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Show the leader names and locations of colleges.
SELECT Leader_Name, College_Location FROM college
CREATE TABLE college (Leader_Name VARCHAR, College_Location VARCHAR)
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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
CREATE TABLE member (Name VARCHAR, College_ID VARCHAR); CREATE TABLE college (Name VARCHAR, College_ID VARCHAR)
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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
CREATE TABLE member (Name VARCHAR, College_ID VARCHAR); CREATE TABLE college (College_Location VARCHAR, College_ID VARCHAR)
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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"
CREATE TABLE college (Leader_Name VARCHAR, College_ID VARCHAR); CREATE TABLE member (College_ID VARCHAR, Country VARCHAR)
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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
CREATE TABLE member (Name VARCHAR, Member_ID VARCHAR); CREATE TABLE round (Decoration_Theme VARCHAR, Member_ID VARCHAR)
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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
CREATE TABLE round (Member_ID VARCHAR, Rank_in_Round INTEGER); CREATE TABLE member (Name VARCHAR, Member_ID VARCHAR)
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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
CREATE TABLE member (Name VARCHAR, Member_ID VARCHAR); CREATE TABLE round (Member_ID VARCHAR)
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List the names of members who did not participate in any round.
SELECT Name FROM member WHERE NOT Member_ID IN (SELECT Member_ID FROM round)
CREATE TABLE member (Name VARCHAR, Member_ID VARCHAR); CREATE TABLE round (Name VARCHAR, Member_ID VARCHAR)
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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
CREATE TABLE documents (document_name VARCHAR, access_count VARCHAR)
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Find the name of the document that has been accessed the greatest number of times, as well as the count of how many times it has been accessed?
SELECT document_name, access_count FROM documents ORDER BY access_count DESC LIMIT 1
CREATE TABLE documents (document_name VARCHAR, access_count VARCHAR)
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Find the types of documents with more than 4 documents.
SELECT document_type_code FROM documents GROUP BY document_type_code HAVING COUNT(*) > 4
CREATE TABLE documents (document_type_code VARCHAR)
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Find the total access count of all documents in the most popular document type.
SELECT SUM(access_count) FROM documents GROUP BY document_type_code ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE documents (access_count INTEGER, document_type_code VARCHAR)
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What is the average access count of documents?
SELECT AVG(access_count) FROM documents
CREATE TABLE documents (access_count INTEGER)
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What is the structure of the document with the least number of accesses?
SELECT t2.document_structure_description FROM documents AS t1 JOIN document_structures AS t2 ON t1.document_structure_code = t2.document_structure_code GROUP BY t1.document_structure_code ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE document_structures (document_structure_description VARCHAR, document_structure_code VARCHAR); CREATE TABLE documents (document_structure_code VARCHAR)
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What is the type of the document named "David CV"?
SELECT document_type_code FROM documents WHERE document_name = "David CV"
CREATE TABLE documents (document_type_code VARCHAR, document_name VARCHAR)
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Find the list of documents that are both in the most three popular type and have the most three popular structure.
SELECT document_name FROM documents GROUP BY document_type_code ORDER BY COUNT(*) DESC LIMIT 3 INTERSECT SELECT document_name FROM documents GROUP BY document_structure_code ORDER BY COUNT(*) DESC LIMIT 3
CREATE TABLE documents (document_name VARCHAR, document_type_code VARCHAR, document_structure_code VARCHAR)
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What document types do have more than 10000 total access number.
SELECT document_type_code FROM documents GROUP BY document_type_code HAVING SUM(access_count) > 10000
CREATE TABLE documents (document_type_code VARCHAR, access_count INTEGER)
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What are all the section titles of the document named "David CV"?
SELECT t2.section_title FROM documents AS t1 JOIN document_sections AS t2 ON t1.document_code = t2.document_code WHERE t1.document_name = "David CV"
CREATE TABLE documents (document_code VARCHAR, document_name VARCHAR); CREATE TABLE document_sections (section_title VARCHAR, document_code VARCHAR)
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Find all the name of documents without any sections.
SELECT document_name FROM documents WHERE NOT document_code IN (SELECT document_code FROM document_sections)
CREATE TABLE document_sections (document_name VARCHAR, document_code VARCHAR); CREATE TABLE documents (document_name VARCHAR, document_code VARCHAR)
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List all the username and passwords of users with the most popular role.
SELECT user_name, password FROM users GROUP BY role_code ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE users (user_name VARCHAR, password VARCHAR, role_code VARCHAR)
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Find the average access counts of documents with functional area "Acknowledgement".
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"
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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Find names of the document without any images.
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
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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What is the name of the document with the most number of sections?
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
CREATE TABLE document_sections (document_code VARCHAR); CREATE TABLE documents (document_name VARCHAR, document_code VARCHAR)
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List all the document names which contains "CV".
SELECT document_name FROM documents WHERE document_name LIKE "%CV%"
CREATE TABLE documents (document_name VARCHAR)
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How many users are logged in?
SELECT COUNT(*) FROM users WHERE user_login = 1
CREATE TABLE users (user_login VARCHAR)
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Find the description of the most popular role among the users that have logged in.
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)
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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Find the average access count of documents with the least popular structure.
SELECT AVG(access_count) FROM documents GROUP BY document_structure_code ORDER BY COUNT(*) LIMIT 1
CREATE TABLE documents (access_count INTEGER, document_structure_code VARCHAR)
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List all the image name and URLs in the order of their names.
SELECT image_name, image_url FROM images ORDER BY image_name
CREATE TABLE images (image_name VARCHAR, image_url VARCHAR)
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Find the number of users in each role.
SELECT COUNT(*), role_code FROM users GROUP BY role_code
CREATE TABLE users (role_code VARCHAR)
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What document types have more than 2 corresponding documents?
SELECT document_type_code FROM documents GROUP BY document_type_code HAVING COUNT(*) > 2
CREATE TABLE documents (document_type_code VARCHAR)
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How many companies are there?
SELECT COUNT(*) FROM Companies
CREATE TABLE Companies (Id VARCHAR)
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List the names of companies in descending order of market value.
SELECT name FROM Companies ORDER BY Market_Value_billion DESC
CREATE TABLE Companies (name VARCHAR, Market_Value_billion VARCHAR)
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What are the names of companies whose headquarters are not "USA"?
SELECT name FROM Companies WHERE Headquarters <> 'USA'
CREATE TABLE Companies (name VARCHAR, Headquarters VARCHAR)
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What are the name and assets of each company, sorted in ascending order of company name?
SELECT name, Assets_billion FROM Companies ORDER BY name
CREATE TABLE Companies (name VARCHAR, Assets_billion VARCHAR)
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What are the average profits of companies?
SELECT AVG(Profits_billion) FROM Companies
CREATE TABLE Companies (Profits_billion INTEGER)
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What are the maximum and minimum sales of the companies whose industries are not "Banking".
SELECT MAX(Sales_billion), MIN(Sales_billion) FROM Companies WHERE Industry <> "Banking"
CREATE TABLE Companies (Sales_billion INTEGER, Industry VARCHAR)
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How many different industries are the companies in?
SELECT COUNT(DISTINCT Industry) FROM Companies
CREATE TABLE Companies (Industry VARCHAR)
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List the names of buildings in descending order of building height.
SELECT name FROM buildings ORDER BY Height DESC
CREATE TABLE buildings (name VARCHAR, Height VARCHAR)
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Find the stories of the building with the largest height.
SELECT Stories FROM buildings ORDER BY Height DESC LIMIT 1
CREATE TABLE buildings (Stories VARCHAR, Height VARCHAR)
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List the name of a building along with the name of a company whose office is in the building.
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
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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Show the names of the buildings that have more than one company offices.
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
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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Show the name of the building that has the most company offices.
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
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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Please show the names of the buildings whose status is "on-hold", in ascending order of stories.
SELECT name FROM buildings WHERE Status = "on-hold" ORDER BY Stories
CREATE TABLE buildings (name VARCHAR, Status VARCHAR, Stories VARCHAR)
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