Read first
Task 25
Aggregate over a join
SUM after a join totals across related tables.
Task
What does GROUP BY students.name with SUM give?
The editor already contains a starting point. Change it until every check passes — the Next task button unlocks when it does.
What you start with
SELECT students.name, SUM(fees.amount) AS paid_total FROM fees JOIN students ON students.id = fees.student_id WHERE fees.paid = 1 GROUP BY students.name;
Syntax reference
SELECT students.name, SUM(fees.amount) AS paid_total FROM fees JOIN students ON students.id = fees.student_id WHERE fees.paid = 1 GROUP BY students.name;
Read it, then write it yourself. Copy-paste teaches nothing.
The sample database
CREATE TABLE courses (id INTEGER PRIMARY KEY, title TEXT, monthly_fee INTEGER); CREATE TABLE students (id INTEGER PRIMARY KEY, name TEXT, city TEXT, course_id INTEGER, joined TEXT); CREATE TABLE fees (id INTEGER PRIMARY KEY, student_id INTEGER, month TEXT, amount INTEGER, paid INTEGER); INSERT INTO courses VALUES (1,'Web Design',9000),(2,'Python',11667),(3,'JavaScript',11667),(4,'AI & Prompt Engineering',15000); INSERT INTO students VALUES (1,'Sara','Faisalabad',2,'2026-01-12'), (2,'Ali','Faisalabad',1,'2026-01-20'), (3,'Hina','Jhumra',2,'2026-02-02'), (4,'Bilal','Faisalabad',4,'2026-02-15'), (5,'Ayesha','Sargodha',3,'2026-03-01'), (6,'Usman','Jhumra',1,'2026-03-11'), (7,'Zainab','Faisalabad',4,'2026-04-05'), (8,'Hamza','Faisalabad',3,'2026-04-19'); INSERT INTO fees VALUES (1,1,'2026-01',11667,1),(2,1,'2026-02',11667,1), (3,2,'2026-01',9000,1),(4,2,'2026-02',9000,0), (5,3,'2026-02',11667,1),(6,3,'2026-03',11667,0), (7,4,'2026-02',15000,1),(8,4,'2026-03',15000,1), (9,5,'2026-03',11667,0), (10,6,'2026-03',9000,1),(11,6,'2026-04',9000,0), (12,7,'2026-04',15000,0), (13,8,'2026-04',11667,1);
SQL
1 automatic check Section 3 · Joins and subqueries
Code editor
Ctrl + Enter runs
Run your code to see the output and the checks.
Your result
Submit your code to see results
Write your code in the middle column and press Run code.
Why this works
Checks on this task:
- Check 1
One correct answer:
SELECT students.name, SUM(fees.amount) AS paid_total FROM fees JOIN students ON students.id = fees.student_id WHERE fees.paid = 1 GROUP BY students.name;
Track progress
0 of 56 tasks complete
Next task
Next task unlocks when every check passes.
Sign in to keep your code and certificates.
Next task (pass every check first)
Hint
Ask yourself: What does GROUP BY students.name with SUM give?
Still stuck? Open “Syntax reference” in the left column, then type it yourself and change one value.