Read first
Task 16
Count distinct
COUNT(DISTINCT column) counts how many different values appear.
Task
COUNT(DISTINCT city) returns…
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 COUNT(DISTINCT city) FROM students;
Syntax reference
SELECT COUNT(DISTINCT city) FROM students;
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 2 · Summaries
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.
Next task (pass every check first)
Hint
Ask yourself: COUNT(DISTINCT city) returns…
Still stuck? Open “Syntax reference” in the left column, then type it yourself and change one value.