EasySQL challenge #71GROUP BY

Users per Country

Count users grouped by country.

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.

Solve This SQL Challenge

The task

Count users grouped by country.

A correct answer returns 8 rows with 2 columns: country, 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
id	first_name	last_name	email	phone	city	country	created_at	status
1	Jennifer	Torres	jennifer.torres57@icloud.com	+1-263-376-2371	Houston	Canada	2024-04-13	active
2	Donna	Adams	donna.adams52@gmail.com	+1-753-780-6670	Fort Worth	Canada	2020-09-15	inactive
3	Kenneth	Young	kenneth.young94@proton.me	+1-440-743-9503	Austin	USA	2023-09-05	active

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
country	COUNT(*)
Australia	14
Brazil	14
Canada	21
France	10
Germany	13
India	10
UK	16
USA	52

The complete result: 8 rows. This is the output of the model solution on the practice database. The task does not fix a sort order, so your rows may come back in a different order.

What you practise

Summarising many rows into one value with COUNT, SUM, AVG, MIN or MAX
Grouping rows with GROUP BY to get one result row per group

Hint and approach

Try the challenge first. Open these only when you need them.

Show the hint

GROUP BY country

Show the step-by-step approach

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

Read from the users table.
Group the rows by country, so the result has one row for each value.
For each group, compute COUNT(*).

Model solution

Checking your session…

What to watch for

GROUP BY. Every selected column must either be in GROUP BY or sit inside an aggregate. SQLite accepts other columns and returns an arbitrary value for them; most databases raise an error.
Aggregate functions. Aggregates skip NULLs. AVG divides by the number of non-NULL values and COUNT(column) can be smaller than COUNT(*).