EasySQL challenge #17GROUP BYCOUNT

Orders by Status

Count orders grouped by status.

The challenge runs on the SQLabHub practice database (SQLite) and uses the orders table. You can solve it in the browser, with nothing to install.

Solve This SQL Challenge

The task

Count orders grouped by status.

A correct answer returns 6 rows with 2 columns: status, COUNT(*).

Tables used in this challenge

`orders` 200 rows

id INTEGER (PK) · user_id INTEGER · total REAL · discount REAL · item_count INTEGER · status TEXT · order_date TEXT · shipped_date TEXT · shipping_city TEXT · shipping_country TEXT

First 3 rows of the orders table
id	user_id	total	discount	item_count	status	order_date	shipped_date	shipping_city	shipping_country
1	43	2638.11	0.25	3	cancelled	2024-04-24	NULL	Los Angeles	USA
2	8	2947	0.19	5	delivered	2024-03-15	2023-04-09	Chicago	USA
3	136	683.24	0.08	2	delivered	2023-02-15	2023-07-10	Los Angeles	USA

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
status	COUNT(*)
cancelled	27
delivered	82
pending	23
processing	28
refunded	19
shipped	21

The complete result: 6 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

Use GROUP BY

Show the step-by-step approach

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

Read from the orders table.
Group the rows by status, 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(*).