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
Show all document names using templates with template type code BK.
SELECT T2.document_name FROM Templates AS T1 JOIN Documents AS T2 ON T1.template_id = T2.template_id WHERE T1.template_type_code = 'BK'
CREATE TABLE Documents (document_name VARCHAR(255), template_id VARCHAR(255)); CREATE TABLE Templates (template_id VARCHAR(255), template_type_code VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 4138 }
Show all template type codes and the number of documents using each type.
SELECT T1.template_type_code, COUNT(*) FROM Templates AS T1 JOIN Documents AS T2 ON T1.template_id = T2.template_id GROUP BY T1.template_type_code
CREATE TABLE Documents (template_id VARCHAR(255), document_name VARCHAR(255)); CREATE TABLE Templates (template_type_code VARCHAR(255), template_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4139 }
Which template type code is used by most number of documents?
SELECT T1.template_type_code FROM Templates AS T1 JOIN Documents AS T2 ON T1.template_id = T2.template_id GROUP BY T1.template_type_code ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Documents (template_id VARCHAR(255), document_name VARCHAR(255)); CREATE TABLE Templates (template_type_code VARCHAR(255), template_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
0
{ "split": "train", "row_index": 4140 }
Show all template type codes and descriptions.
SELECT template_type_code, template_type_description FROM Ref_template_types
CREATE TABLE Ref_template_types (template_type_code VARCHAR(255), template_type_description VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 4142 }
What is the template type descriptions for template type code "AD".
SELECT template_type_description FROM Ref_template_types WHERE template_type_code = 'AD'
CREATE TABLE Ref_template_types (template_type_description VARCHAR(255), template_type_code VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 4143 }
What is the template type code for template type description "Book".
SELECT template_type_code FROM Ref_template_types WHERE template_type_description = 'Book'
CREATE TABLE Ref_template_types (template_type_code VARCHAR(255), template_type_description VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 4144 }
What are the distinct template type descriptions for the templates ever used by any document?
SELECT DISTINCT T1.template_type_description FROM Ref_template_types AS T1 JOIN Templates AS T2 ON T1.template_type_code = T2.template_type_code JOIN Documents AS T3 ON T2.Template_ID = T3.template_ID
CREATE TABLE Templates (template_type_code VARCHAR(255), Template_ID VARCHAR(255)); CREATE TABLE Documents (template_ID VARCHAR(255), document_name VARCHAR(255)); CREATE TABLE Ref_template_types (template_type_description VARCHAR(255), template_type_code VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 4145 }
What are the template ids with template type description "Presentation".
SELECT T2.template_id FROM Ref_template_types AS T1 JOIN Templates AS T2 ON T1.template_type_code = T2.template_type_code WHERE T1.template_type_description = 'Presentation'
CREATE TABLE Templates (template_id VARCHAR(255), template_type_code VARCHAR(255)); CREATE TABLE Ref_template_types (template_type_code VARCHAR(255), template_type_description VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 4146 }
How many paragraphs in total?
SELECT COUNT(*) FROM Paragraphs
CREATE TABLE Paragraphs (Id VARCHAR(255), document_ID VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 4147 }
How many paragraphs for the document with name 'Summer Show'?
SELECT COUNT(*) FROM Paragraphs AS T1 JOIN Documents AS T2 ON T1.document_ID = T2.document_ID WHERE T2.document_name = 'Summer Show'
CREATE TABLE Documents (document_ID VARCHAR(255), document_name VARCHAR(255)); CREATE TABLE Paragraphs (document_ID VARCHAR(255), paragraph_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 4148 }
Show paragraph details for paragraph with text 'Korea ' .
SELECT other_details FROM paragraphs WHERE paragraph_text LIKE 'korea'
CREATE TABLE paragraphs (other_details VARCHAR(255), paragraph_text VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 4149 }
Show all paragraph ids and texts for the document with name 'Welcome to NY'.
SELECT T1.paragraph_id, T1.paragraph_text FROM Paragraphs AS T1 JOIN Documents AS T2 ON T1.document_id = T2.document_id WHERE T2.Document_Name = 'Welcome to NY'
CREATE TABLE Documents (document_id VARCHAR(255), Document_Name VARCHAR(255)); CREATE TABLE Paragraphs (paragraph_id VARCHAR(255), paragraph_text VARCHAR(255), document_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4150 }
Show all paragraph texts for the document "Customer reviews".
SELECT T1.paragraph_text FROM Paragraphs AS T1 JOIN Documents AS T2 ON T1.document_id = T2.document_id WHERE T2.document_name = 'Customer reviews'
CREATE TABLE Paragraphs (paragraph_text VARCHAR(255), document_id VARCHAR(255)); CREATE TABLE Documents (document_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": 4151 }
Show all document ids and the number of paragraphs in each document. Order by document id.
SELECT document_id, COUNT(*) FROM Paragraphs GROUP BY document_id ORDER BY document_id
CREATE TABLE Paragraphs (document_id VARCHAR(255), paragraph_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4152 }
Show all document ids, names and the number of paragraphs in each document.
SELECT T1.document_id, T2.document_name, COUNT(*) FROM Paragraphs AS T1 JOIN Documents AS T2 ON T1.document_id = T2.document_id GROUP BY T1.document_id
CREATE TABLE Documents (document_name VARCHAR(255), document_id VARCHAR(255)); CREATE TABLE Paragraphs (document_id VARCHAR(255), paragraph_id VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4153 }
List all document ids with at least two paragraphs.
SELECT document_id FROM Paragraphs GROUP BY document_id HAVING COUNT(*) >= 2
CREATE TABLE Paragraphs (document_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": 4154 }
What is the document id and name with greatest number of paragraphs?
SELECT T1.document_id, T2.document_name FROM Paragraphs AS T1 JOIN Documents AS T2 ON T1.document_id = T2.document_id GROUP BY T1.document_id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Documents (document_name VARCHAR(255), document_id VARCHAR(255)); CREATE TABLE Paragraphs (document_id VARCHAR(255), id INTEGER PRIMARY KEY AUTO_INCREMENT)
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 4155 }
What is the document id with least number of paragraphs?
SELECT document_id FROM Paragraphs GROUP BY document_id ORDER BY COUNT(*) LIMIT 1
CREATE TABLE Paragraphs (document_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": 4156 }
What is the document id with 1 to 2 paragraphs?
SELECT document_id FROM Paragraphs GROUP BY document_id HAVING COUNT(*) BETWEEN 1 AND 2
CREATE TABLE Paragraphs (document_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": 4157 }
Show the document id with paragraph text 'Brazil' and 'Ireland'.
SELECT document_id FROM Paragraphs WHERE paragraph_text = 'Brazil' INTERSECT SELECT document_id FROM Paragraphs WHERE paragraph_text = 'Ireland'
CREATE TABLE Paragraphs (document_id VARCHAR(255), paragraph_text VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 4158 }
How many teachers are there?
SELECT COUNT(*) FROM teacher
CREATE TABLE teacher (Id VARCHAR(255), Name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 4159 }
List the names of teachers in ascending order of age.
SELECT Name FROM teacher ORDER BY Age
CREATE TABLE teacher (Name VARCHAR(255), Age VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 4160 }
What are the age and hometown of teachers?
SELECT Age, Hometown FROM teacher
CREATE TABLE teacher (Age VARCHAR(255), Hometown VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 4161 }
List the name of teachers whose hometown is not `` Little Lever Urban District '' .
SELECT name FROM teacher WHERE hometown <> 'little lever urban district'
CREATE TABLE teacher (name VARCHAR(255), hometown VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 4162 }
Show the name of teachers aged either 32 or 33?
SELECT Name FROM teacher WHERE Age = 32 OR Age = 33
CREATE TABLE teacher (Name VARCHAR(255), Age VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4163 }
What is the hometown of the youngest teacher?
SELECT Hometown FROM teacher ORDER BY Age LIMIT 1
CREATE TABLE teacher (Hometown VARCHAR(255), Age VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 4164 }
Show different hometown of teachers and the number of teachers from each hometown.
SELECT Hometown, COUNT(*) FROM teacher GROUP BY Hometown
CREATE TABLE teacher (Hometown 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": 4165 }
List the most common hometown of teachers.
SELECT Hometown FROM teacher GROUP BY Hometown ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE teacher (Hometown 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": 4166 }
Show the hometowns shared by at least two teachers.
SELECT Hometown FROM teacher GROUP BY Hometown HAVING COUNT(*) >= 2
CREATE TABLE teacher (Hometown 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": 4167 }
Show names of teachers and the courses they are arranged to teach.
SELECT T3.Name, T2.Course FROM course_arrange AS T1 JOIN course AS T2 ON T1.Course_ID = T2.Course_ID JOIN teacher AS T3 ON T1.Teacher_ID = T3.Teacher_ID
CREATE TABLE course_arrange (Course_ID VARCHAR(255), Teacher_ID VARCHAR(255)); CREATE TABLE teacher (Name VARCHAR(255), Teacher_ID VARCHAR(255)); CREATE TABLE course (Course VARCHAR(255), Course_ID VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4168 }
Show names of teachers and the courses they are arranged to teach in ascending alphabetical order of the teacher's name.
SELECT T3.Name, T2.Course FROM course_arrange AS T1 JOIN course AS T2 ON T1.Course_ID = T2.Course_ID JOIN teacher AS T3 ON T1.Teacher_ID = T3.Teacher_ID ORDER BY T3.Name
CREATE TABLE course_arrange (Course_ID VARCHAR(255), Teacher_ID VARCHAR(255)); CREATE TABLE teacher (Name VARCHAR(255), Teacher_ID VARCHAR(255)); CREATE TABLE course (Course VARCHAR(255), Course_ID VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 4169 }
Show the name of the teacher for the math course.
SELECT T3.Name FROM course_arrange AS T1 JOIN course AS T2 ON T1.Course_ID = T2.Course_ID JOIN teacher AS T3 ON T1.Teacher_ID = T3.Teacher_ID WHERE T2.Course = 'Math'
CREATE TABLE course_arrange (Course_ID VARCHAR(255), Teacher_ID VARCHAR(255)); CREATE TABLE course (Course_ID VARCHAR(255), Course VARCHAR(255)); CREATE TABLE teacher (Name VARCHAR(255), Teacher_ID VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 4170 }
Show names of teachers and the number of courses they teach.
SELECT T2.Name, COUNT(*) FROM course_arrange AS T1 JOIN teacher AS T2 ON T1.Teacher_ID = T2.Teacher_ID GROUP BY T2.Name
CREATE TABLE teacher (Name VARCHAR(255), Teacher_ID VARCHAR(255)); CREATE TABLE course_arrange (Teacher_ID VARCHAR(255), Course_ID VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4171 }
Show names of teachers that teach at least two courses.
SELECT T2.Name FROM course_arrange AS T1 JOIN teacher AS T2 ON T1.Teacher_ID = T2.Teacher_ID GROUP BY T2.Name HAVING COUNT(*) >= 2
CREATE TABLE teacher (Name VARCHAR(255), Teacher_ID VARCHAR(255)); CREATE TABLE course_arrange (Teacher_ID VARCHAR(255), Course_ID VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 4172 }
List the names of teachers who have not been arranged to teach courses.
SELECT Name FROM teacher WHERE NOT Teacher_id IN (SELECT Teacher_id FROM course_arrange)
CREATE TABLE course_arrange (Name VARCHAR(255), Teacher_id VARCHAR(255)); CREATE TABLE teacher (Name VARCHAR(255), Teacher_id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 4173 }
How many visitors below age 30 are there?
SELECT COUNT(*) FROM visitor WHERE age < 30
CREATE TABLE visitor (age 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": 4174 }
Find the names of the visitors whose membership level is higher than 4, and order the results by the level from high to low.
SELECT name FROM visitor WHERE Level_of_membership > 4 ORDER BY Level_of_membership DESC
CREATE TABLE visitor (name VARCHAR(255), Level_of_membership INTEGER)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 4175 }
What is the average age of the visitors whose membership level is not higher than 4?
SELECT AVG(age) FROM visitor WHERE Level_of_membership <= 4
CREATE TABLE visitor (age INTEGER, Level_of_membership VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 4176 }
Find the name and membership level of the visitors whose membership level is higher than 4, and sort by their age from old to young.
SELECT name, Level_of_membership FROM visitor WHERE Level_of_membership > 4 ORDER BY age DESC
CREATE TABLE visitor (name VARCHAR(255), Level_of_membership INTEGER, age VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4177 }
Find the id and name of the museum that has the most staff members?
SELECT museum_id, name FROM museum ORDER BY num_of_staff DESC LIMIT 1
CREATE TABLE museum (museum_id VARCHAR(255), name VARCHAR(255), num_of_staff VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4178 }
What are the opening year and staff number of the museum named Plaza Museum?
SELECT Num_of_Staff, Open_Year FROM museum WHERE name = 'Plaza Museum'
CREATE TABLE museum (Num_of_Staff VARCHAR(255), Open_Year VARCHAR(255), name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 4180 }
find the names of museums which have more staff than the minimum staff number of all museums opened after 2010.
SELECT name FROM museum WHERE num_of_staff > (SELECT MIN(num_of_staff) FROM museum WHERE open_year > 2010)
CREATE TABLE museum (name VARCHAR(255), num_of_staff INTEGER, open_year INTEGER)
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 4181 }
find the id, name and age for visitors who visited some museums more than once.
SELECT t1.id, t1.name, t1.age FROM visitor AS t1 JOIN visit AS t2 ON t1.id = t2.visitor_id GROUP BY t1.id HAVING COUNT(*) > 1
CREATE TABLE visit (visitor_id VARCHAR(255), visit_date VARCHAR(255)); CREATE TABLE visitor (id VARCHAR(255), name VARCHAR(255), age VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4182 }
What are the id, name and membership level of visitors who have spent the largest amount of money in total in all museum tickets?
SELECT t2.visitor_id, t1.name, t1.Level_of_membership FROM visitor AS t1 JOIN visit AS t2 ON t1.id = t2.visitor_id GROUP BY t2.visitor_id ORDER BY SUM(t2.Total_spent) DESC LIMIT 1
CREATE TABLE visit (visitor_id VARCHAR(255), Total_spent INTEGER); CREATE TABLE visitor (name VARCHAR(255), Level_of_membership VARCHAR(255), id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 4183 }
What are the id and name of the museum visited most times?
SELECT t2.Museum_ID, t1.name FROM museum AS t1 JOIN visit AS t2 ON t1.Museum_ID = t2.Museum_ID GROUP BY t2.Museum_ID ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE museum (name VARCHAR(255), Museum_ID VARCHAR(255)); CREATE TABLE visit (Museum_ID VARCHAR(255), visit_date VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 4184 }
Find the name and age of the visitor who bought the most tickets at once.
SELECT t1.name, t1.age FROM visitor AS t1 JOIN visit AS t2 ON t1.id = t2.visitor_id ORDER BY t2.num_of_ticket DESC LIMIT 1
CREATE TABLE visitor (name VARCHAR(255), age VARCHAR(255), id VARCHAR(255)); CREATE TABLE visit (visitor_id VARCHAR(255), num_of_ticket VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 4186 }
What are the average and maximum number of tickets bought in all visits?
SELECT AVG(num_of_ticket), MAX(num_of_ticket) FROM visit
CREATE TABLE visit (num_of_ticket 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": 4187 }
What is the total ticket expense of the visitors whose membership level is 1?
SELECT SUM(t2.Total_spent) FROM visitor AS t1 JOIN visit AS t2 ON t1.id = t2.visitor_id WHERE t1.Level_of_membership = 1
CREATE TABLE visit (Total_spent INTEGER, visitor_id VARCHAR(255)); CREATE TABLE visitor (id VARCHAR(255), Level_of_membership VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 4188 }
What is the name of the visitor who visited both a museum opened before 2009 and a museum opened after 2011?
SELECT t1.name FROM visitor AS t1 JOIN visit AS t2 ON t1.id = t2.visitor_id JOIN museum AS t3 ON t3.Museum_ID = t2.Museum_ID WHERE t3.open_year < 2009 INTERSECT SELECT t1.name FROM visitor AS t1 JOIN visit AS t2 ON t1.id = t2.visitor_id JOIN museum AS t3 ON t3.Museum_ID = t2.Museum_ID WHERE t3.open_year > 2011
CREATE TABLE visitor (name VARCHAR(255), id VARCHAR(255)); CREATE TABLE museum (Museum_ID VARCHAR(255), open_year INTEGER); CREATE TABLE visit (visitor_id VARCHAR(255), Museum_ID VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
1
0
{ "split": "train", "row_index": 4189 }
Find the number of visitors who did not visit any museum opened after 2010.
SELECT COUNT(*) FROM visitor WHERE NOT id IN (SELECT t2.visitor_id FROM museum AS t1 JOIN visit AS t2 ON t1.Museum_ID = t2.Museum_ID WHERE t1.open_year > 2010)
CREATE TABLE museum (Museum_ID VARCHAR(255), open_year INTEGER); CREATE TABLE visitor (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE visit (visitor_id VARCHAR(255), Museum_ID VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
1
{ "split": "train", "row_index": 4190 }
How many museums were opened after 2013 or before 2008?
SELECT COUNT(*) FROM museum WHERE open_year > 2013 OR open_year < 2008
CREATE TABLE museum (open_year 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": 4191 }
Find the total number of players.
SELECT COUNT(*) FROM players
CREATE TABLE players (Id VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 4192 }
Find the total number of matches.
SELECT COUNT(*) FROM matches
CREATE TABLE matches (Id VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 4193 }
List the first name and birth date of all players from the country with code USA.
SELECT first_name, birth_date FROM players WHERE country_code = 'USA'
CREATE TABLE players (first_name VARCHAR(255), birth_date VARCHAR(255), country_code VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 4194 }
find the number of distinct country codes of all players.
SELECT COUNT(DISTINCT country_code) FROM players
CREATE TABLE players (country_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": 4198 }
Find the number of distinct name of losers.
SELECT COUNT(DISTINCT loser_name) FROM matches
CREATE TABLE matches (loser_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": 4199 }
Find the name of tourney that has more than 10 matches.
SELECT tourney_name FROM matches GROUP BY tourney_name HAVING COUNT(*) > 10
CREATE TABLE matches (tourney_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": 4200 }
List the names of all winners who played in both 2013 and 2016.
SELECT winner_name FROM matches WHERE YEAR = 2013 INTERSECT SELECT winner_name FROM matches WHERE YEAR = 2016
CREATE TABLE matches (winner_name VARCHAR(255), YEAR VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
0
{ "split": "train", "row_index": 4201 }
List the number of all matches who played in years of 2013 or 2016.
SELECT COUNT(*) FROM matches WHERE YEAR = 2013 OR YEAR = 2016
CREATE TABLE matches (YEAR 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": 4202 }
What are the country code and first name of the players who won in both tourney WTA Championships and Australian Open?
SELECT T1.country_code, T1.first_name FROM players AS T1 JOIN matches AS T2 ON T1.player_id = T2.winner_id WHERE T2.tourney_name = 'WTA Championships' INTERSECT SELECT T1.country_code, T1.first_name FROM players AS T1 JOIN matches AS T2 ON T1.player_id = T2.winner_id WHERE T2.tourney_name = 'Australian Open'
CREATE TABLE matches (winner_id VARCHAR(255), tourney_name VARCHAR(255)); CREATE TABLE players (country_code VARCHAR(255), first_name VARCHAR(255), player_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
1
1
{ "split": "train", "row_index": 4203 }
Find the first name and country code of the oldest player.
SELECT first_name, country_code FROM players ORDER BY birth_date LIMIT 1
CREATE TABLE players (first_name VARCHAR(255), country_code VARCHAR(255), birth_date VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4204 }
List the first and last name of all players in the order of birth date.
SELECT first_name, last_name FROM players ORDER BY birth_date
CREATE TABLE players (first_name VARCHAR(255), last_name VARCHAR(255), birth_date VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 4205 }
List the first and last name of all players who are left / L hand in the order of birth date.
SELECT first_name, last_name FROM players WHERE hand = 'L' ORDER BY birth_date
CREATE TABLE players (first_name VARCHAR(255), last_name VARCHAR(255), hand VARCHAR(255), birth_date VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4206 }
Find the first name and country code of the player who did the most number of tours.
SELECT T1.country_code, T1.first_name FROM players AS T1 JOIN rankings AS T2 ON T1.player_id = T2.player_id ORDER BY T2.tours DESC LIMIT 1
CREATE TABLE players (country_code VARCHAR(255), first_name VARCHAR(255), player_id VARCHAR(255)); CREATE TABLE rankings (player_id VARCHAR(255), tours VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 4207 }
Find the year that has the most number of matches.
SELECT YEAR FROM matches GROUP BY YEAR ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE matches (YEAR 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": 4208 }
Find the name and rank points of the winner who won the most times.
SELECT winner_name, winner_rank_points FROM matches GROUP BY winner_name ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE matches (winner_name VARCHAR(255), winner_rank_points VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 4209 }
Find the name of the winner who has the highest rank points and participated in the Australian Open tourney.
SELECT winner_name FROM matches WHERE tourney_name = 'Australian Open' ORDER BY winner_rank_points DESC LIMIT 1
CREATE TABLE matches (winner_name VARCHAR(255), tourney_name VARCHAR(255), winner_rank_points VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
0
{ "split": "train", "row_index": 4210 }
find the names of loser and winner who played in the match with greatest number of minutes.
SELECT winner_name, loser_name FROM matches ORDER BY minutes DESC LIMIT 1
CREATE TABLE matches (winner_name VARCHAR(255), loser_name VARCHAR(255), minutes VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4211 }
find the number of players for each country.
SELECT COUNT(*), country_code FROM players GROUP BY country_code
CREATE TABLE players (country_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": 4214 }
find the code of the country where has the greatest number of players.
SELECT country_code FROM players GROUP BY country_code ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE players (country_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": 4215 }
Find the codes of countries that have more than 50 players.
SELECT country_code FROM players GROUP BY country_code HAVING COUNT(*) > 50
CREATE TABLE players (country_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": 4216 }
Find the total number of tours for each ranking date.
SELECT SUM(tours), ranking_date FROM rankings GROUP BY ranking_date
CREATE TABLE rankings (ranking_date VARCHAR(255), tours INTEGER)
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 4217 }
Find the number of matches happened in each year.
SELECT COUNT(*), YEAR FROM matches GROUP BY YEAR
CREATE TABLE matches (YEAR 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": 4218 }
Find the name and rank of the 3 youngest winners across all matches.
SELECT DISTINCT winner_name, winner_rank FROM matches ORDER BY winner_age LIMIT 3
CREATE TABLE matches (winner_name VARCHAR(255), winner_rank VARCHAR(255), winner_age VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4219 }
How many different winners both participated in the WTA Championships and were left handed?
SELECT COUNT(DISTINCT winner_name) FROM matches WHERE tourney_name = 'WTA Championships' AND winner_hand = 'L'
CREATE TABLE matches (winner_name VARCHAR(255), tourney_name VARCHAR(255), winner_hand VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 4220 }
Find the first name, country code and birth date of the winner who has the highest rank points in all matches.
SELECT T1.first_name, T1.country_code, T1.birth_date FROM players AS T1 JOIN matches AS T2 ON T1.player_id = T2.winner_id ORDER BY T2.winner_rank_points DESC LIMIT 1
CREATE TABLE players (first_name VARCHAR(255), country_code VARCHAR(255), birth_date VARCHAR(255), player_id VARCHAR(255)); CREATE TABLE matches (winner_id VARCHAR(255), winner_rank_points VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 4221 }
Find the number of players for each hand type.
SELECT COUNT(*), hand FROM players GROUP BY hand
CREATE TABLE players (hand 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": 4222 }
How many ships ended up being 'Captured'?
SELECT COUNT(*) FROM ship WHERE disposition_of_ship = 'Captured'
CREATE TABLE ship (disposition_of_ship 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": 4223 }
List the name and tonnage ordered by in descending alphaetical order for the names.
SELECT name, tonnage FROM ship ORDER BY name DESC
CREATE TABLE ship (name VARCHAR(255), tonnage VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 4224 }
List the name, date and result of each battle.
SELECT name, date FROM battle
CREATE TABLE battle (name VARCHAR(255), date VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 4225 }
What is maximum and minimum death toll caused each time?
SELECT MAX(killed), MIN(killed) FROM death
CREATE TABLE death (killed 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": 4226 }
What is the average number of injuries caused each time?
SELECT AVG(injured) FROM death
CREATE TABLE death (injured 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": 4227 }
What are the name and results of the battles when the bulgarian commander is not 'Boril'
SELECT name, RESULT FROM battle WHERE bulgarian_commander <> 'Boril'
CREATE TABLE battle (name VARCHAR(255), RESULT VARCHAR(255), bulgarian_commander VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 4229 }
What are the different ids and names of the battles that lost any 'Brig' type shipes?
SELECT DISTINCT T1.id, T1.name FROM battle AS T1 JOIN ship AS T2 ON T1.id = T2.lost_in_battle WHERE T2.ship_type = 'Brig'
CREATE TABLE ship (lost_in_battle VARCHAR(255), ship_type VARCHAR(255)); CREATE TABLE battle (id VARCHAR(255), name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
1
{ "split": "train", "row_index": 4230 }
What are the ids and names of the battles that led to more than 10 people killed in total.
SELECT T1.id, T1.name FROM battle AS T1 JOIN ship AS T2 ON T1.id = T2.lost_in_battle JOIN death AS T3 ON T2.id = T3.caused_by_ship_id GROUP BY T1.id HAVING SUM(T3.killed) > 10
CREATE TABLE death (caused_by_ship_id VARCHAR(255), killed INTEGER); CREATE TABLE battle (id VARCHAR(255), name VARCHAR(255)); CREATE TABLE ship (lost_in_battle VARCHAR(255), id VARCHAR(255))
generic
hard
hard
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
3
0
1
{ "split": "train", "row_index": 4231 }
What are the distinct battle names which are between bulgarian commander 'Kaloyan' and latin commander 'Baldwin I'?
SELECT name FROM battle WHERE bulgarian_commander = 'Kaloyan' AND latin_commander = 'Baldwin I'
CREATE TABLE battle (name VARCHAR(255), bulgarian_commander VARCHAR(255), latin_commander VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
1
{ "split": "train", "row_index": 4233 }
How many different results are there for the battles?
SELECT COUNT(DISTINCT RESULT) FROM battle
CREATE TABLE battle (RESULT 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": 4234 }
How many battles did not lose any ship with tonnage '225'?
SELECT COUNT(*) FROM battle WHERE NOT id IN (SELECT lost_in_battle FROM ship WHERE tonnage = '225')
CREATE TABLE ship (id VARCHAR(255), lost_in_battle VARCHAR(255), tonnage VARCHAR(255)); CREATE TABLE battle (id VARCHAR(255), lost_in_battle VARCHAR(255), tonnage VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
1
1
{ "split": "train", "row_index": 4235 }
List the name and date the battle that has lost the ship named 'Lettice' and the ship named 'HMS Atalanta'
SELECT T1.name, T1.date FROM battle AS T1 JOIN ship AS T2 ON T1.id = T2.lost_in_battle WHERE T2.name = 'Lettice' INTERSECT SELECT T1.name, T1.date FROM battle AS T1 JOIN ship AS T2 ON T1.id = T2.lost_in_battle WHERE T2.name = 'HMS Atalanta'
CREATE TABLE ship (lost_in_battle VARCHAR(255), name VARCHAR(255)); CREATE TABLE battle (name VARCHAR(255), date VARCHAR(255), id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
1
1
{ "split": "train", "row_index": 4236 }
What are the notes of the death events which has substring 'East'?
SELECT note FROM death WHERE note LIKE '%East%'
CREATE TABLE death (note 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": 4238 }
what are all the addresses including line 1 and line 2?
SELECT line_1, line_2 FROM addresses
CREATE TABLE addresses (line_1 VARCHAR(255), line_2 VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 4239 }
How many courses in total are listed?
SELECT COUNT(*) FROM Courses
CREATE TABLE Courses (Id VARCHAR(255), name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 4240 }
How is the math course described?
SELECT course_description FROM Courses WHERE course_name = 'math'
CREATE TABLE Courses (course_description VARCHAR(255), course_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 4241 }
What is the zip code of the address in the city Port Chelsea?
SELECT zip_postcode FROM Addresses WHERE city = 'Port Chelsea'
CREATE TABLE Addresses (zip_postcode VARCHAR(255), city VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
1
0
0
{ "split": "train", "row_index": 4242 }
Which department offers the most number of degrees? List department name and id.
SELECT T2.department_name, T1.department_id FROM Degree_Programs AS T1 JOIN Departments AS T2 ON T1.department_id = T2.department_id GROUP BY T1.department_id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE Degree_Programs (department_id VARCHAR(255), program_name VARCHAR(255)); CREATE TABLE Departments (department_name VARCHAR(255), department_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 4243 }
What is the name and id of the department with the most number of degrees ?
SELECT t2.department_name, t1.department_id FROM degree_programs AS t1 JOIN departments AS t2 ON t1.department_id = t2.department_id GROUP BY t1.department_id ORDER BY COUNT(*) DESC LIMIT 1
CREATE TABLE degree_programs (department_id VARCHAR(255), program_name VARCHAR(255)); CREATE TABLE departments (department_name VARCHAR(255), department_id VARCHAR(255))
generic
extra
extra
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
4
0
1
{ "split": "train", "row_index": 4244 }
How many departments offer any degree?
SELECT COUNT(DISTINCT department_id) FROM Degree_Programs
CREATE TABLE Degree_Programs (department_id VARCHAR(255), program_name VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 4245 }
How many different degree names are offered?
SELECT COUNT(DISTINCT degree_summary_name) FROM Degree_Programs
CREATE TABLE Degree_Programs (degree_summary_name VARCHAR(255), degree_program_id VARCHAR(255))
generic
easy
easy
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
0
{ "split": "train", "row_index": 4246 }
How many degrees does the engineering department offer?
SELECT COUNT(*) FROM Departments AS T1 JOIN Degree_Programs AS T2 ON T1.department_id = T2.department_id WHERE T1.department_name = 'engineer'
CREATE TABLE Degree_Programs (department_id VARCHAR(255), degree_name VARCHAR(255)); CREATE TABLE Departments (department_id VARCHAR(255), department_name VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
2
0
0
{ "split": "train", "row_index": 4247 }
What are the names and descriptions of all the sections?
SELECT section_name, section_description FROM Sections
CREATE TABLE Sections (section_name VARCHAR(255), section_description VARCHAR(255))
generic
medium
medium
classified
valid
valid
valid
[]
[]
[]
[]
[ "validated", "extended", "classified" ]
null
0
0
1
{ "split": "train", "row_index": 4248 }