HardSQL challenge #330GROUP BYAGGREGATIONANALYSIS

Find Each Customer's Busiest Day by Order Count

Return each user with the highest number of orders they placed on a single day.

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

Find users who placed the highest number of orders in a single day.

A correct answer returns 114 rows with 2 columns: user_id, max_orders.

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	max_orders
150	1
149	1
148	1
147	1
146	1

First 5 of 114 rows. 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
Grouping rows with GROUP BY to get one result row per group
Grouping by more than one column to get a row per combination
Querying the result of another query with a subquery in FROM
Sorting the result with ORDER BY

Hint and approach

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

Show the hint

GROUP BY user_id, order_date + MAX

Show the step-by-step approach

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

Build an inner query first (a derived table):
1. Read from the orders table.
2. Group the rows by user_id and order_date, so the result has one row for each combination.
3. For each group, compute COUNT(*).
Group the rows by user_id, so the result has one row for each value.
For each group, compute MAX(cnt).
Sort the result by max_orders descending.

Model solution

Checking your session…

What to watch for

MAX(COUNT(*)) is not valid SQL. The inner query counts orders per user and day, and the outer query takes the maximum of those counts per user.

Derived table. Columns computed in the inner query are only visible outside under their alias, so every calculated column there needs a name.
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.
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.

Solve This SQL Challenge