sharkiefff/RBAC-Text2SQL-Benchmark huggingface dataset:
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: What are the ids of students who both have friends and are liked?
Answer: SELECT student_id FROM Friend INTERSECT SELECT liked_id FROM Likes
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT student_id FROM Friend INTERSECT SELECT liked_id FROM Likes
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: Show name of all students who have some friends and also are liked by someone else.
Answer: SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id INTERSECT SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.liked_id  =  T2.id
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id INTERSECT SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.liked_id  =  T2.id
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: What are the names of high schoolers who both have friends and are liked?
Answer: SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id INTERSECT SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.liked_id  =  T2.id
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id INTERSECT SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.liked_id  =  T2.id
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: Count the number of likes for each student id.
Answer: SELECT student_id ,  count(*) FROM Likes GROUP BY student_id
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT student_id ,  count(*) FROM Likes GROUP BY student_id
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: How many likes correspond to each student id?
Answer: SELECT student_id ,  count(*) FROM Likes GROUP BY student_id
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT student_id ,  count(*) FROM Likes GROUP BY student_id
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: Show the names of high schoolers who have likes, and numbers of likes for each.
Answer: SELECT T2.name ,  count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id GROUP BY T1.student_id
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T2.name ,  count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id GROUP BY T1.student_id
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: What are the names of high schoolers who have likes, and how many likes does each have?
Answer: SELECT T2.name ,  count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id GROUP BY T1.student_id
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T2.name ,  count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id GROUP BY T1.student_id
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: What is the name of the high schooler who has the greatest number of likes?
Answer: SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id GROUP BY T1.student_id ORDER BY count(*) DESC LIMIT 1
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id GROUP BY T1.student_id ORDER BY count(*) DESC LIMIT 1
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: Give the name of the student with the most likes.
Answer: SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id GROUP BY T1.student_id ORDER BY count(*) DESC LIMIT 1
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id GROUP BY T1.student_id ORDER BY count(*) DESC LIMIT 1
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: Show the names of students who have at least 2 likes.
Answer: SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id GROUP BY T1.student_id HAVING count(*)  >=  2
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id GROUP BY T1.student_id HAVING count(*)  >=  2
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: What are the names of students who have 2 or more likes?
Answer: SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id GROUP BY T1.student_id HAVING count(*)  >=  2
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id GROUP BY T1.student_id HAVING count(*)  >=  2
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: Show the names of students who have a grade higher than 5 and have at least 2 friends.
Answer: SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id WHERE T2.grade  >  5 GROUP BY T1.student_id HAVING count(*)  >=  2
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id WHERE T2.grade  >  5 GROUP BY T1.student_id HAVING count(*)  >=  2
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: What are the names of high schoolers who have a grade of over 5 and have 2 or more friends?
Answer: SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id WHERE T2.grade  >  5 GROUP BY T1.student_id HAVING count(*)  >=  2
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id WHERE T2.grade  >  5 GROUP BY T1.student_id HAVING count(*)  >=  2
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: How many likes does Kyle have?
Answer: SELECT count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id WHERE T2.name  =  "Kyle"
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id WHERE T2.name  =  "Kyle"
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: Return the number of likes that the high schooler named Kyle has.
Answer: SELECT count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id WHERE T2.name  =  "Kyle"
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id WHERE T2.name  =  "Kyle"
##Instruction:
Database: network_1
Total Tables: 3
Total Columns: 7

Table Schemas:

network_1 contains tables such as Highschooler, Friend, Likes.
Table Highschooler has columns such as ID, name, grade.
ID is the primary key.
Table Friend has columns such as student_id, friend_id.
student_id is the primary key.
Table Likes has columns such as student_id, liked_id.
student_id is the primary key.
The friend_id of Friend is the foreign key of ID of Highschooler.
The student_id of Friend is the foreign key of ID of Highschooler.
The student_id of Likes is the foreign key of ID of Highschooler.
The liked_id of Likes is the foreign key of ID of Highschooler.

##Role Access Policy (Column-Level):
Role: StudentCounselor
Accessible Columns: Highschooler: ID, name, grade; Friend: student_id, friend_id; Likes: student_id, liked_id

Based on the above schema and your role's column-level access policy, generate a SQL query to answer the question. If your role does not have access to the required columns, respond with: "Sorry, I cannot answer."


Question: Find the average grade of all students who have some friends.
Answer: SELECT avg(grade) FROM Highschooler WHERE id IN (SELECT T1.student_id FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id)
Permission: allowed (All required columns are permitted)
Gold SQL: SELECT avg(grade) FROM Highschooler WHERE id IN (SELECT T1.student_id FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id  =  T2.id)