MediumSQL challenge #142SUBQUERY

High Discount Orders

Find orders where discount is above average.

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 orders where discount is above average.

A correct answer returns 96 rows with 10 columns: id, user_id, total, discount, item_count, status, order_date, shipped_date, shipping_city, shipping_country.

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
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
5	140	1120.99	0.13	4	delivered	2023-06-16	2023-12-21	Denver	USA
7	69	1755.54	0.2	2	refunded	2023-03-22	NULL	Philadelphia	USA
8	91	4163.77	0.2	2	delivered	2022-01-17	2023-08-22	Indianapolis	USA

First 5 of 96 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

Returning every column of a table with SELECT *
Filtering rows with a WHERE condition
Summarising many rows into one value with COUNT, SUM, AVG, MIN or MAX
Using the result of one query inside another

Hint and approach

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

Show the hint

Subquery AVG

Show the step-by-step approach

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

Read from the orders table.
Filter in WHERE: keep the rows where discount is greater than the value returned by a subquery.
The subquery runs once and returns a single value. It works like this:
1. Read from the orders table.
2. Aggregate over all the rows that are left: compute AVG(discount). With no GROUP BY, the result is a single row.
Return every column with SELECT *.

Model solution

Checking your session…

What to watch for

Subquery. A subquery compared with =, > or < must return exactly one row and one column. If it can return several rows, use IN or an aggregate.
Aggregate functions. Aggregates skip NULLs. AVG divides by the number of non-NULL values and COUNT(column) can be smaller than COUNT(*).
WHERE. WHERE is evaluated before the SELECT list, so it cannot use a column alias defined there or the result of an aggregate.