Count the Rows in a Table with COUNT(*)
Return the total number of users as a single number.
The challenge runs on the SQLabHub practice database (SQLite) and uses the users table. You can solve it in the browser, with nothing to install.
The task
Count total number of users.
A correct answer returns 1 row with one column: COUNT(*).
Tables used in this challenge
users 150 rows
id INTEGER (PK) · first_name TEXT · last_name TEXT · email TEXT · phone TEXT · city TEXT · country TEXT · created_at TEXT · status TEXT · is_premium INTEGER
| id | first_name | last_name | phone | city | country | created_at | status | is_premium | |
|---|---|---|---|---|---|---|---|---|---|
The first rows of each table are shown. See how all the tables relate on the database schema page.
Expected output
| COUNT(*) |
|---|
The complete result: 1 row. This is the output of the model solution on the practice database.
What you practise
- Summarising many rows into one value with COUNT, SUM, AVG, MIN or MAX
Hint and approach
Try the challenge first. Open these only when you need them.
Show the hint
Use COUNT(*)
Show the step-by-step approach
How the model solution is built, in the order the database evaluates it:
- Read from the
userstable. - Aggregate over all the rows that are left: compute
COUNT(*). With no GROUP BY, the result is a single row.
Model solution
Checking your session…
What to watch for
COUNT(*) counts rows. COUNT(column) counts only the rows where that column is not NULL, so the two can disagree on a table with missing values.
- Aggregate functions. Aggregates skip NULLs. AVG divides by the number of non-NULL values and COUNT(column) can be smaller than COUNT(*).
Related SQL challenges
- Find the Lowest Price with MIN()Similar levelEasy
- Find the Highest Price with MAX()Similar levelEasy
- Find the Last Order Date for Each CustomerHarderMedium
- Calculate the Average Price with AVG()Similar levelEasy
- Add Up Order Totals with SUM()Similar levelEasy
Same technique, other tables
More practice on this pattern with different columns:
Keep learning
- Lesson: Aggregate Functions
- All SQL practice challenges, by difficulty and topic
- SQL interview questions with answers
- SQL Playground: run any query on the same database