MediumSQL challenge #165HAVINGSUBQUERY

Find Customers with an Above-Average Number of Orders

Return the users whose order count is higher than the average order count per user.

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.

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The task

Find users whose order count is above average.

A correct answer returns 59 rows with 2 columns: user_id, cnt.

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	cnt
4	3
6	2
7	3
8	3
10	2

First 5 of 59 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
Filtering groups after aggregation with HAVING
Using the result of one query inside another
Querying the result of another query with a subquery in FROM

Hint and approach

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

Show the hint

HAVING + subquery

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, so the result has one row for each value.
For each group, compute COUNT(*).
Filter the groups in HAVING: keep those where cnt is greater than the value returned by a subquery.
The subquery in HAVING runs once and returns a single value. It works like this:
1. Build an inner query first (a derived table):
  1. Read from the orders table.
  2. Group the rows by user_id, so the result has one row for each value.
  3. For each group, compute COUNT(*).
2. Aggregate over all the rows that are left: compute AVG(order_count). With no GROUP BY, the result is a single row.

Model solution

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What to watch for

AVG(COUNT(*)) is not allowed, so the average is taken over a derived table that first counts orders per user. Users with no orders are not part of that average, because they have no rows in orders.

Derived table. Columns computed in the inner query are only visible outside under their alias, so every calculated column there needs a name.
HAVING. Conditions on single rows belong in WHERE and conditions on aggregates in HAVING. Putting a row filter in HAVING works, but it groups rows that are about to be thrown away.
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.

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