MediumSQL challenge #145GROUP BY

Count Orders per Customer per Day with a Two-Column GROUP BY

Return the number of orders for each combination of user and order date.

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 per user per day.

A correct answer returns 200 rows with 3 columns: user_id, order_date, 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
user_id	order_date	COUNT(*)
1	2022-09-26	1
2	2023-04-02	1
4	2022-08-22	1
4	2024-04-20	1
4	2024-11-15	1

First 5 of 200 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 by more than one column to get a row per combination

Hint and approach

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

Show the hint

GROUP BY user_id, date

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 user_id and order_date, so the result has one row for each combination.
For each group, compute COUNT(*).

Model solution

Checking your session…

What to watch for

Grouping by two columns makes one group per distinct pair. In this dataset no customer ordered twice on the same day, so every count is 1; the pattern is the standard first step for finding repeat activity within a day.

GROUP BY on several columns. Each extra GROUP BY column splits the groups further. Grouping by a column that is unique per row gives back one group per row.
Aggregate functions. Aggregates skip NULLs. AVG divides by the number of non-NULL values and COUNT(column) can be smaller than COUNT(*).