question
string
answer
string
context
string
dialect
string
spider_hardness
string
extended_hardness
string
classification_status
string
syntax_status
string
ddl_syntax_status
string
schema_status
string
syntax_errors
list
ddl_syntax_errors
list
schema_errors
list
classification_errors
list
status_history
list
correction_status
string
component1
int64
component2
int64
others
int64
validation_source
dict
List every individual's first name, middle name and last name in alphabetical order by last name.
SELECT individual_first_name, individual_middle_name, individual_last_name FROM individuals ORDER BY individual_last_name
CREATE TABLE individuals (individual_first_name VARCHAR(255), individual_middle_name VARCHAR(255), individual_last_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3578 }
List all the types of forms.
SELECT DISTINCT form_type_code FROM forms
CREATE TABLE forms (form_type_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3579 }
Find the name of the most popular party form.
SELECT t1.form_name FROM forms AS t1 JOIN party_forms AS t2 ON t1.form_id = t2.form_id GROUP BY t2.form_id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE party_forms (form_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE forms (form_name VARCHAR(255), form_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
0
{ "split": "train", "row_index": 3580 }
Find the payment method and phone of the party with email "enrico09@example.com".
SELECT payment_method_code, party_phone FROM parties WHERE party_email = 'enrico09@example.com'
CREATE TABLE parties (payment_method_code VARCHAR(255), party_phone VARCHAR(255), party_email VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3581 }
List all the name of organizations in order of the date formed.
SELECT organization_name FROM organizations ORDER BY date_formed
CREATE TABLE organizations (organization_name VARCHAR(255), date_formed VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3583 }
Find the name of the youngest organization.
SELECT organization_name FROM organizations ORDER BY date_formed DESC LIMIT 1
CREATE TABLE organizations (organization_name VARCHAR(255), date_formed VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 3584 }
Find the last name of the latest contact individual of the organization "Labour Party".
SELECT t3.individual_last_name FROM organizations AS t1 JOIN organization_contact_individuals AS t2 ON t1.organization_id = t2.organization_id JOIN individuals AS t3 ON t2.individual_id = t3.individual_id WHERE t1.organization_name = 'Labour Party' ORDER BY t2.date_contact_to DESC LIMIT 1
CREATE TABLE organizations (organization_id VARCHAR(255), organization_name VARCHAR(255)); CREATE TABLE individuals (individual_last_name VARCHAR(255), individual_id VARCHAR(255)); CREATE TABLE organization_contact_individuals (organization_id VARCHAR(255), individual_id VARCHAR(255), date_contact_to VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
5
0
0
{ "split": "train", "row_index": 3585 }
Find the name of all the cities and states.
SELECT town_city FROM addresses UNION SELECT state_province_county FROM addresses
CREATE TABLE addresses (town_city VARCHAR(255), state_province_county VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
1
0
{ "split": "train", "row_index": 3588 }
How many cities are there in state "Colorado"?
SELECT COUNT(*) FROM addresses WHERE state_province_county = 'Colorado'
CREATE TABLE addresses (state_province_county VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3589 }
Find the payment method code used by more than 3 parties.
SELECT payment_method_code FROM parties GROUP BY payment_method_code HAVING COUNT(*) > 3
CREATE TABLE parties (payment_method_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3590 }
Find the name of organizations whose names contain "Party".
SELECT organization_name FROM organizations WHERE organization_name LIKE '%Party%'
CREATE TABLE organizations (organization_name VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 3591 }
How many distinct payment methods are used by parties?
SELECT COUNT(DISTINCT payment_method_code) FROM parties
CREATE TABLE parties (payment_method_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3592 }
Which is the email of the party that has used the services the most number of times?
SELECT t1.party_email FROM parties AS t1 JOIN party_services AS t2 ON t1.party_id = t2.customer_id GROUP BY t1.party_email ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE parties (party_email VARCHAR(255), party_id VARCHAR(255)); CREATE TABLE party_services (customer_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
0
{ "split": "train", "row_index": 3593 }
Which state can address "6862 Kaitlyn Knolls" possibly be in?
SELECT state_province_county FROM addresses WHERE line_1_number_building LIKE '%6862 Kaitlyn Knolls%'
CREATE TABLE addresses (state_province_county VARCHAR(255), line_1_number_building VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 3594 }
What is the name of organization that has the greatest number of contact individuals?
SELECT t1.organization_name FROM organizations AS t1 JOIN organization_contact_individuals AS t2 ON t1.organization_id = t2.organization_id GROUP BY t1.organization_name ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE organization_contact_individuals (organization_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE organizations (organization_name VARCHAR(255), organization_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
0
{ "split": "train", "row_index": 3595 }
Find the last name of the individuals that have been contact individuals of an organization.
SELECT DISTINCT t1.individual_last_name FROM individuals AS t1 JOIN organization_contact_individuals AS t2 ON t1.individual_id = t2.individual_id
CREATE TABLE individuals (individual_last_name VARCHAR(255), individual_id VARCHAR(255)); CREATE TABLE organization_contact_individuals (individual_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3596 }
How many drivers are there?
SELECT COUNT(*) FROM driver
CREATE TABLE driver (Id VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3597 }
Show the name, home city, and age for all drivers.
SELECT name, home_city, age FROM driver
CREATE TABLE driver (name VARCHAR(255), home_city VARCHAR(255), age VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 3598 }
Show the party and the number of drivers in each party.
SELECT party, COUNT(*) FROM driver GROUP BY party
CREATE TABLE driver (party VARCHAR(255), name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3599 }
Show the name of drivers in descending order of age.
SELECT name FROM driver ORDER BY age DESC
CREATE TABLE driver (name VARCHAR(255), age VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3600 }
Show all different home cities.
SELECT DISTINCT home_city FROM driver
CREATE TABLE driver (home_city VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3601 }
Show the home city with the most number of drivers.
SELECT home_city FROM driver GROUP BY home_city ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE driver (home_city VARCHAR(255), name VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 3602 }
Show the party with drivers from Hartford and drivers older than 40.
SELECT party FROM driver WHERE home_city = 'Hartford' AND age > 40
CREATE TABLE driver (party VARCHAR(255), home_city VARCHAR(255), age VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3603 }
Show home city where at least two drivers older than 40 are from.
SELECT home_city FROM driver WHERE age > 40 GROUP BY home_city HAVING COUNT(*) >= 2
CREATE TABLE driver (home_city VARCHAR(255), age INTEGER)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 3604 }
Show all home cities except for those having a driver older than 40.
SELECT home_city FROM driver EXCEPT SELECT home_city FROM driver WHERE age > 40
CREATE TABLE driver (home_city VARCHAR(255), age INTEGER)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
1
0
{ "split": "train", "row_index": 3605 }
Show the names of the drivers without a school bus.
SELECT name FROM driver WHERE NOT driver_id IN (SELECT driver_id FROM school_bus)
CREATE TABLE school_bus (name VARCHAR(255), driver_id VARCHAR(255)); CREATE TABLE driver (name VARCHAR(255), driver_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 3606 }
Show the types of schools that have two schools.
SELECT TYPE FROM school GROUP BY TYPE HAVING COUNT(*) = 2
CREATE TABLE school (TYPE VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3607 }
Show the school name and driver name for all school buses.
SELECT T2.school, T3.name FROM school_bus AS T1 JOIN school AS T2 ON T1.school_id = T2.school_id JOIN driver AS T3 ON T1.driver_id = T3.driver_id
CREATE TABLE school (school VARCHAR(255), school_id VARCHAR(255)); CREATE TABLE school_bus (school_id VARCHAR(255), driver_id VARCHAR(255)); CREATE TABLE driver (name VARCHAR(255), driver_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 3608 }
What is the maximum, minimum and average years spent working on a school bus?
SELECT MAX(years_working), MIN(years_working), AVG(years_working) FROM school_bus
CREATE TABLE school_bus (years_working INTEGER, id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
2
{ "split": "train", "row_index": 3609 }
Show the school name and type for schools without a school bus.
SELECT school, TYPE FROM school WHERE NOT school_id IN (SELECT school_id FROM school_bus)
CREATE TABLE school_bus (school VARCHAR(255), TYPE VARCHAR(255), school_id VARCHAR(255)); CREATE TABLE school (school VARCHAR(255), TYPE VARCHAR(255), school_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
1
{ "split": "train", "row_index": 3610 }
Show the type of school and the number of buses for each type.
SELECT T2.type, COUNT(*) FROM school_bus AS T1 JOIN school AS T2 ON T1.school_id = T2.school_id GROUP BY T2.type
CREATE TABLE school (type VARCHAR(255), school_id VARCHAR(255)); CREATE TABLE school_bus (school_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 3611 }
How many drivers are from Hartford city or younger than 40?
SELECT COUNT(*) FROM driver WHERE home_city = 'Hartford' OR age < 40
CREATE TABLE driver (home_city VARCHAR(255), age VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 3612 }
List names for drivers from Hartford city and younger than 40.
SELECT name FROM driver WHERE home_city = 'Hartford' AND age < 40
CREATE TABLE driver (name VARCHAR(255), home_city VARCHAR(255), age VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3613 }
How many flights have a velocity larger than 200?
SELECT COUNT(*) FROM flight WHERE velocity > 200
CREATE TABLE flight (velocity INTEGER, id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3615 }
List the vehicle flight number, date and pilot of all the flights, ordered by altitude.
SELECT vehicle_flight_number, date, pilot FROM flight ORDER BY altitude
CREATE TABLE flight (vehicle_flight_number VARCHAR(255), date VARCHAR(255), pilot VARCHAR(255), altitude VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3616 }
List the id, country, city and name of the airports ordered alphabetically by the name.
SELECT id, country, city, name FROM airport ORDER BY name
CREATE TABLE airport (id VARCHAR(255), country VARCHAR(255), city VARCHAR(255), name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3617 }
What is maximum group equity shareholding of the companies?
SELECT MAX(group_equity_shareholding) FROM operate_company
CREATE TABLE operate_company (group_equity_shareholding INTEGER, id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3618 }
What is the velocity of the pilot named 'Thompson'?
SELECT AVG(velocity) FROM flight WHERE pilot = 'Thompson'
CREATE TABLE flight (velocity INTEGER, pilot VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3619 }
What are the names of the airports which are not in the country 'Iceland'?
SELECT name FROM airport WHERE country <> 'Iceland'
CREATE TABLE airport (name VARCHAR(255), country VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3621 }
What is the id, name and IATA code of the airport that had most number of flights?
SELECT T1.id, T1.name, T1.IATA FROM airport AS T1 JOIN flight AS T2 ON T1.id = T2.airport_id GROUP BY T2.id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE airport (id VARCHAR(255), name VARCHAR(255), IATA VARCHAR(255)); CREATE TABLE flight (id VARCHAR(255), airport_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 3624 }
What are the different pilot names who had piloted a flight in the country 'United States' or in the airport named 'Billund Airport'?
SELECT DISTINCT T2.pilot FROM airport AS T1 JOIN flight AS T2 ON T1.id = T2.airport_id WHERE T1.country = 'United States' OR T1.name = 'Billund Airport'
CREATE TABLE airport (id VARCHAR(255), country VARCHAR(255), name VARCHAR(255)); CREATE TABLE flight (pilot VARCHAR(255), airport_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 3625 }
What is the most common company type, and how many are there?
SELECT TYPE, COUNT(*) FROM operate_company GROUP BY TYPE ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE operate_company (TYPE VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
2
{ "split": "train", "row_index": 3626 }
How many airports haven't the pilot 'Thompson' driven an aircraft?
SELECT COUNT(*) FROM airport WHERE NOT id IN (SELECT airport_id FROM flight WHERE pilot = 'Thompson')
CREATE TABLE airport (id VARCHAR(255), airport_id VARCHAR(255), pilot VARCHAR(255)); CREATE TABLE flight (id VARCHAR(255), airport_id VARCHAR(255), pilot VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
1
{ "split": "train", "row_index": 3627 }
Which of the airport names contains the word 'international'?
SELECT name FROM airport WHERE name LIKE '%international%'
CREATE TABLE airport (name VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 3629 }
how many airports are there in each country?
SELECT COUNT(*), country FROM airport GROUP BY country
CREATE TABLE airport (country VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3631 }
which countries have more than 2 airports?
SELECT country FROM airport GROUP BY country HAVING COUNT(*) > 2
CREATE TABLE airport (country VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3632 }
which pilot is in charge of the most number of flights?
SELECT pilot FROM flight GROUP BY pilot ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE flight (pilot VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 3633 }
Show all account ids and account details.
SELECT account_id, account_details FROM Accounts
CREATE TABLE Accounts (account_id VARCHAR(255), account_details VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 3634 }
How many statements do we have?
SELECT COUNT(*) FROM Statements
CREATE TABLE Statements (Id VARCHAR(255), info VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3635 }
List all statement ids and statement details.
SELECT STATEMENT_ID, statement_details FROM Statements
CREATE TABLE Statements (STATEMENT_ID VARCHAR(255), statement_details VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 3636 }
Show statement id, statement detail, account detail for accounts.
SELECT T1.statement_id, T2.statement_details, T1.account_details FROM Accounts AS T1 JOIN Statements AS T2 ON T1.statement_id = T2.statement_id
CREATE TABLE Accounts (statement_id VARCHAR(255), account_details VARCHAR(255)); CREATE TABLE Statements (statement_details VARCHAR(255), statement_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3637 }
Show all statement id and the number of accounts for each statement.
SELECT STATEMENT_ID, COUNT(*) FROM Accounts GROUP BY STATEMENT_ID
CREATE TABLE Accounts (STATEMENT_ID VARCHAR(255), account_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3638 }
Show the statement id and the statement detail for the statement with most number of accounts.
SELECT T1.statement_id, T2.statement_details FROM Accounts AS T1 JOIN Statements AS T2 ON T1.statement_id = T2.statement_id GROUP BY T1.statement_id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Accounts (statement_id VARCHAR(255), account_name VARCHAR(255)); CREATE TABLE Statements (statement_details VARCHAR(255), statement_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 3639 }
Show the number of documents.
SELECT COUNT(*) FROM Documents
CREATE TABLE Documents (Id VARCHAR(255), title VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3640 }
List the document type code, document name, and document description for the document with name 'Noel CV' or name 'King Book'.
SELECT document_type_code, document_name, document_description FROM Documents WHERE document_name = 'Noel CV' OR document_name = 'King Book'
CREATE TABLE Documents (document_type_code VARCHAR(255), document_name VARCHAR(255), document_description VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
2
{ "split": "train", "row_index": 3641 }
Show the ids and names of all documents.
SELECT document_id, document_name FROM Documents
CREATE TABLE Documents (document_id VARCHAR(255), document_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 3642 }
Find names and ids of all documents with document type code BK.
SELECT document_name, document_id FROM Documents WHERE document_type_code = 'BK'
CREATE TABLE Documents (document_name VARCHAR(255), document_id VARCHAR(255), document_type_code VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3643 }
How many documents are with document type code BK for each product id?
SELECT COUNT(*), project_id FROM Documents WHERE document_type_code = 'BK' GROUP BY project_id
CREATE TABLE Documents (project_id VARCHAR(255), document_type_code VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 3644 }
Show project ids and the number of documents in each project.
SELECT project_id, COUNT(*) FROM Documents GROUP BY project_id
CREATE TABLE Documents (project_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3646 }
What is the id of the project with least number of documents?
SELECT project_id FROM Documents GROUP BY project_id ORDER BY COUNT(*) LIMIT 1
CREATE TABLE Documents (project_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 3647 }
Show the ids for projects with at least 2 documents.
SELECT project_id FROM Documents GROUP BY project_id HAVING COUNT(*) >= 2
CREATE TABLE Documents (project_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3648 }
List document type codes and the number of documents in each code.
SELECT document_type_code, COUNT(*) FROM Documents GROUP BY document_type_code
CREATE TABLE Documents (document_type_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3649 }
What is the document type code with most number of documents?
SELECT document_type_code FROM Documents GROUP BY document_type_code ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Documents (document_type_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 3650 }
Show the document type code with fewer than 3 documents.
SELECT document_type_code FROM Documents GROUP BY document_type_code HAVING COUNT(*) < 3
CREATE TABLE Documents (document_type_code VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3651 }
Show the statement detail and the corresponding document name for the statement with detail 'Private Project'.
SELECT T1.statement_details, T2.document_name FROM Statements AS T1 JOIN Documents AS T2 ON T1.statement_id = T2.document_id WHERE T1.statement_details = 'Private Project'
CREATE TABLE Statements (statement_details VARCHAR(255), statement_id VARCHAR(255)); CREATE TABLE Documents (document_name VARCHAR(255), document_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 3652 }
Show all document type codes, document type names, document type descriptions.
SELECT document_type_code, document_type_name, document_type_description FROM Ref_document_types
CREATE TABLE Ref_document_types (document_type_code VARCHAR(255), document_type_name VARCHAR(255), document_type_description VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 3653 }
What is the document type description for document type named Film?
SELECT document_type_description FROM Ref_document_types WHERE document_type_name = 'Film'
CREATE TABLE Ref_document_types (document_type_description VARCHAR(255), document_type_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3654 }
What is the document type name and the document type description and creation date for all the documents?
SELECT T1.document_type_name, T1.document_type_description, T2.Document_date FROM Ref_document_types AS T1 JOIN Documents AS T2 ON T1.document_type_code = T2.document_type_code
CREATE TABLE Ref_document_types (document_type_name VARCHAR(255), document_type_description VARCHAR(255), document_type_code VARCHAR(255)); CREATE TABLE Documents (Document_date VARCHAR(255), document_type_code VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3655 }
Show the number of projects.
SELECT COUNT(*) FROM Projects
CREATE TABLE Projects (Id VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3656 }
List ids and details for all projects.
SELECT project_id, project_details FROM Projects
CREATE TABLE Projects (project_id VARCHAR(255), project_details VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 3657 }
What is the project id and detail for the project with at least two documents?
SELECT T1.project_id, T1.project_details FROM Projects AS T1 JOIN Documents AS T2 ON T1.project_id = T2.project_id GROUP BY T1.project_id HAVING COUNT(*) > 2
CREATE TABLE Projects (project_id VARCHAR(255), project_details VARCHAR(255)); CREATE TABLE Documents (project_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 3658 }
What is the project detail for the project with document "King Book"?
SELECT T1.project_details FROM Projects AS T1 JOIN Documents AS T2 ON T1.project_id = T2.project_id WHERE T2.document_name = 'King Book'
CREATE TABLE Projects (project_details VARCHAR(255), project_id VARCHAR(255)); CREATE TABLE Documents (project_id VARCHAR(255), document_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 3659 }
How many budget types do we have?
SELECT COUNT(*) FROM Ref_budget_codes
CREATE TABLE Ref_budget_codes (Id VARCHAR(255), description VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3660 }
List all budget type codes and descriptions.
SELECT budget_type_code, budget_type_description FROM Ref_budget_codes
CREATE TABLE Ref_budget_codes (budget_type_code VARCHAR(255), budget_type_description VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 3661 }
What is the description for the budget type with code ORG?
SELECT budget_type_description FROM Ref_budget_codes WHERE budget_type_code = 'ORG'
CREATE TABLE Ref_budget_codes (budget_type_description VARCHAR(255), budget_type_code VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3662 }
How many documents have expenses?
SELECT COUNT(*) FROM Documents_with_expenses
CREATE TABLE Documents_with_expenses (Id VARCHAR(255), budget_type_code VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3663 }
What are the document ids for the budget type code 'SF'?
SELECT document_id FROM Documents_with_expenses WHERE budget_type_code = 'SF'
CREATE TABLE Documents_with_expenses (document_id VARCHAR(255), budget_type_code VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3664 }
Show the budget type code and description and the corresponding document id.
SELECT T2.budget_type_code, T2.budget_type_description, T1.document_id FROM Documents_with_expenses AS T1 JOIN Ref_budget_codes AS T2 ON T1.budget_type_code = T2.budget_type_code
CREATE TABLE Ref_budget_codes (budget_type_code VARCHAR(255), budget_type_description VARCHAR(255)); CREATE TABLE Documents_with_expenses (document_id VARCHAR(255), budget_type_code VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3665 }
Show ids for all documents with budget types described as 'Government'.
SELECT T1.document_id FROM Documents_with_expenses AS T1 JOIN Ref_Budget_Codes AS T2 ON T1.Budget_Type_code = T2.Budget_Type_code WHERE T2.budget_type_Description = 'Government'
CREATE TABLE Documents_with_expenses (document_id VARCHAR(255), Budget_Type_code VARCHAR(255)); CREATE TABLE Ref_Budget_Codes (Budget_Type_code VARCHAR(255), budget_type_Description VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 3666 }
Show budget type codes and the number of documents in each budget type.
SELECT budget_type_code, COUNT(*) FROM Documents_with_expenses GROUP BY budget_type_code
CREATE TABLE Documents_with_expenses (budget_type_code VARCHAR(255), document_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 3667 }
What is the budget type code with most number of documents.
SELECT budget_type_code FROM Documents_with_expenses GROUP BY budget_type_code ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Documents_with_expenses (budget_type_code VARCHAR(255), document_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 3668 }
What are the ids of documents which don't have expense budgets?
SELECT document_id FROM Documents EXCEPT SELECT document_id FROM Documents_with_expenses
CREATE TABLE Documents (document_id VARCHAR(255), document_name VARCHAR(255)); CREATE TABLE Documents_with_expenses (document_id VARCHAR(255), budget_type_code VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
1
0
{ "split": "train", "row_index": 3669 }
Show ids for all documents in type CV without expense budgets.
SELECT document_id FROM Documents WHERE document_type_code = 'CV' EXCEPT SELECT document_id FROM Documents_with_expenses
CREATE TABLE Documents_with_expenses (document_id VARCHAR(255), document_type_code VARCHAR(255)); CREATE TABLE Documents (document_id VARCHAR(255), document_type_code VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 3670 }
What are the ids of documents with letter 's' in the name with any expense budgets.
SELECT T1.document_id FROM Documents AS T1 JOIN Documents_with_expenses AS T2 ON T1.document_id = T2.document_id WHERE T1.document_name LIKE '%s%'
CREATE TABLE Documents_with_expenses (document_id VARCHAR(255), budget_type_code VARCHAR(255)); CREATE TABLE Documents (document_id VARCHAR(255), document_name VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 3671 }
How many documents do not have any expense?
SELECT COUNT(*) FROM Documents WHERE NOT document_id IN (SELECT document_id FROM Documents_with_expenses)
CREATE TABLE Documents (document_id VARCHAR(255), document_type_code VARCHAR(255)); CREATE TABLE Documents_with_expenses (document_id VARCHAR(255), budget_type_code VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
1
{ "split": "train", "row_index": 3672 }
What are the dates for the documents with both 'GV' type and 'SF' type expenses?
SELECT T1.document_date FROM Documents AS T1 JOIN Documents_with_Expenses AS T2 ON T1.document_id = T2.document_id WHERE T2.budget_type_code = 'GV' INTERSECT SELECT T1.document_date FROM Documents AS T1 JOIN Documents_with_Expenses AS T2 ON T1.document_id = T2.document_id WHERE T2.budget_type_code = 'SF'
CREATE TABLE Documents_with_Expenses (document_id VARCHAR(255), budget_type_code VARCHAR(255)); CREATE TABLE Documents (document_date VARCHAR(255), document_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
1
0
{ "split": "train", "row_index": 3673 }
What are the account details with the largest value or with value having char '5' in it?
SELECT MAX(Account_details) FROM Accounts UNION SELECT Account_details FROM Accounts WHERE Account_details LIKE '%5%'
CREATE TABLE Accounts (Account_details INTEGER, id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
1
0
{ "split": "train", "row_index": 3674 }
Find the total number of scientists.
SELECT COUNT(*) FROM scientists
CREATE TABLE scientists (Id VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3675 }
Find the total hours of all projects.
SELECT SUM(hours) FROM projects
CREATE TABLE projects (hours INTEGER, id INTEGER PRIMARY KEY AUTO_INCREMENT, project_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3676 }
How many different scientists are assigned to any project?
SELECT COUNT(DISTINCT scientist) FROM assignedto
CREATE TABLE assignedto (scientist VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3677 }
Find the number of distinct projects.
SELECT COUNT(DISTINCT name) FROM projects
CREATE TABLE projects (name VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3678 }
Find the average hours of all projects.
SELECT AVG(hours) FROM projects
CREATE TABLE projects (hours INTEGER, id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 3679 }
Find the name of project that continues for the longest time.
SELECT name FROM projects ORDER BY hours DESC LIMIT 1
CREATE TABLE projects (name VARCHAR(255), hours VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 3680 }
List the name of all projects that are operated longer than the average working hours of all projects.
SELECT name FROM projects WHERE hours > (SELECT AVG(hours) FROM projects)
CREATE TABLE projects (name VARCHAR(255), hours INTEGER, id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 3681 }
Find the name and hours of project that has the most number of scientists.
SELECT T1.name, T1.hours FROM projects AS T1 JOIN assignedto AS T2 ON T1.code = T2.project GROUP BY T2.project ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE assignedto (project VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE projects (name VARCHAR(255), hours VARCHAR(255), code VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 3682 }
Find the name of the project for which a scientist whose name contains ‘Smith’ is assigned to.
SELECT T2.name FROM assignedto AS T1 JOIN projects AS T2 ON T1.project = T2.code JOIN scientists AS T3 ON T1.scientist = T3.SSN WHERE T3.name LIKE '%Smith%'
CREATE TABLE scientists (SSN VARCHAR(255), name VARCHAR(255)); CREATE TABLE assignedto (project VARCHAR(255), scientist VARCHAR(255)); CREATE TABLE projects (name VARCHAR(255), code VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
0
{ "split": "train", "row_index": 3683 }
Find the total hours of the projects that scientists named Michael Rogers or Carol Smith are assigned to.
SELECT SUM(T2.hours) FROM assignedto AS T1 JOIN projects AS T2 ON T1.project = T2.code JOIN scientists AS T3 ON T1.scientist = T3.SSN WHERE T3.name = 'Michael Rogers' OR T3.name = 'Carol Smith'
CREATE TABLE scientists (SSN VARCHAR(255), name VARCHAR(255)); CREATE TABLE projects (hours INTEGER, code VARCHAR(255)); CREATE TABLE assignedto (project VARCHAR(255), scientist VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 3684 }
Find the name of projects that require between 100 and 300 hours of work.
SELECT name FROM projects WHERE hours BETWEEN 100 AND 300
CREATE TABLE projects (name VARCHAR(255), hours INTEGER, id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3685 }
Find the name of the scientist who worked on both a project named 'Matter of Time' and a project named 'A Puzzling Parallax'.
SELECT T3.name FROM assignedto AS T1 JOIN projects AS T2 ON T1.project = T2.code JOIN scientists AS T3 ON T1.scientist = T3.SSN WHERE T2.name = 'Matter of Time' INTERSECT SELECT T3.name FROM assignedto AS T1 JOIN projects AS T2 ON T1.project = T2.code JOIN scientists AS T3 ON T1.scientist = T3.SSN WHERE T2.name = 'A Pu...
CREATE TABLE projects (code VARCHAR(255), name VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT); CREATE TABLE scientists (name VARCHAR(255), SSN VARCHAR(255)); CREATE TABLE assignedto (project VARCHAR(255), scientist VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
1
0
{ "split": "train", "row_index": 3686 }
List the names of all scientists sorted in alphabetical order.
SELECT name FROM scientists ORDER BY name
CREATE TABLE scientists (name VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 3687 }