HardSQL challenge #201WINDOW FUNCTIONPARTITIONRANKING

Find Each Customer's Largest Order

Return the single highest-value order of each user, with all of its columns.

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 highest order value for each user.

A correct answer returns 114 rows with 11 columns: id, user_id, total, discount, item_count, status, order_date, shipped_date, shipping_city, shipping_country, rn.

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	rn
109	1	4064.75	0.04	3	delivered	2022-09-26	2023-09-19	Austin	USA	1
13	2	1618.29	0.23	3	delivered	2023-04-02	2024-04-02	Seattle	USA	1
87	4	2501.23	0	6	delivered	2022-08-22	2024-09-16	San Francisco	USA	1
127	5	4924.23	0.22	1	delivered	2024-05-18	2024-03-21	Phoenix	USA	1
130	6	4151.02	0.13	4	shipped	2022-01-23	2024-10-27	Indianapolis	USA	1

First 5 of 114 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

Filtering rows with a WHERE condition
Querying the result of another query with a subquery in FROM
Numbering rows with ROW_NUMBER() to pick one or a few per group
Restarting a window calculation for each group with PARTITION BY

Hint and approach

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

Show the hint

ROW_NUMBER partition

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. Number the rows with ROW_NUMBER(), ordered by total descending, restarting for each user_id.
Filter in WHERE: keep the rows where rn equals 1. The filter sits in the outer query because a window function cannot be tested in the WHERE clause of the query that computes it.

Model solution

Checking your session…

What to watch for

GROUP BY with MAX(total) gives the amount but loses the other columns of that order. Numbering each customer's orders by total and keeping number 1 returns the whole row.

ROW_NUMBER(). ROW_NUMBER() breaks ties arbitrarily. When the ORDER BY inside OVER does not identify one row, the row that gets number 1 can change between runs.
PARTITION BY. PARTITION BY is not GROUP BY: it keeps every row and only decides which rows a window function looks at.
Derived table. Columns computed in the inner query are only visible outside under their alias, so every calculated column there needs a name.

Solve This SQL Challenge

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Find Each Customer's Largest Order in Every MonthSimilar levelHard
Find the Top 3 Most Expensive Products in Each CategorySimilar levelHard

Same technique, other tables

More practice on this pattern with different columns:

Keep learning

Lesson: ROW_NUMBER()
Lesson: Window Functions
Lesson: SQL Subqueries
Lesson: WHERE Clause
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