EasySQL challenge #46COUNT

Count Rows That Match a Condition

Return how many users are premium, 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.

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The task

Count number of premium 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

First 3 rows of the users table
idfirst_namelast_nameemailphonecitycountrycreated_atstatusis_premium
1JenniferTorresjennifer.torres57@icloud.com+1-263-376-2371HoustonCanada2024-04-13active0
2DonnaAdamsdonna.adams52@gmail.com+1-753-780-6670Fort WorthCanada2020-09-15inactive0
3KennethYoungkenneth.young94@proton.me+1-440-743-9503AustinUSA2023-09-05active0

The first rows of each table are shown. See how all the tables relate on the database schema page.

Expected output

Expected output of the challenge
COUNT(*)
69

The complete result: 1 row. This is the output of the model solution on the practice database.

What you practise

  • Filtering rows with a WHERE condition
  • 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

COUNT with condition

Show the step-by-step approach

How the model solution is built, in the order the database evaluates it:

  1. Read from the users table.
  2. Filter in WHERE: keep the rows where is_premium equals 1.
  3. Aggregate over all the rows that are left: compute COUNT(*). With no GROUP BY, the result is a single row.

Model solution

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What to watch for

The WHERE clause removes rows before COUNT(*) sees them. Because the flag is 0 or 1, SUM(is_premium) gives the same number, and that form lets you count several conditions in one query.

  • Aggregate functions. Aggregates skip NULLs. AVG divides by the number of non-NULL values and COUNT(column) can be smaller than COUNT(*).
  • WHERE. WHERE is evaluated before the SELECT list, so it cannot use a column alias defined there or the result of an aggregate.
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