HardSQL challenge #310SELF JOINDATEAGGREGATION

Count Each Customer's Orders in a Rolling 30-Day Window

For every order, count the same user's orders in the 30 days that follow, busiest windows first.

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 with maximum number of orders in any 30-day rolling window.

A correct answer returns 200 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
44	2
47	2
124	2
1	1
2	1

First 5 of 200 rows. This is the output of the model solution on the practice database.

What you practise

Filtering a range of values with BETWEEN
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
Combining rows from two tables on a matching key with an inner join
Joining a table to itself under two aliases
Working with dates in SQLite: strftime, julianday and DATE
Sorting the result with ORDER BY

Hint and approach

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

Show the hint

Use self join with date range

Show the step-by-step approach

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

Read from the orders table, under the alias o1.
Join orders again, under the alias o2, on o1.user_id = o2.user_id, and only where o2.order_date is between o1.order_date and 30 days after o1.order_date, both ends included.
Group the rows by o1.user_id and o1.order_date, so the result has one row for each combination.
For each group, compute COUNT(*).
Sort the result by cnt descending.

Model solution

Checking your session…

What to watch for

SQLite window frames cannot span a number of days, so the window is built with a self join on a date range. Each order is its own window start, and the count includes that order itself.

Self join. Without a condition such as a.id < b.id, a self join pairs every row with itself and lists each pair twice.
INNER JOIN. An inner join drops rows that have no partner on the other side, and it repeats a row once for every match, which inflates SUM and COUNT.
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.

Solve This SQL Challenge

Measure Monthly Revenue Volatility per ProductSimilar levelHard
Count Orders in Each Customer's First 7 DaysSimilar levelHard
Count Orders per Customer per Day with a Two-Column GROUP BYEasierMedium
Find Each Customer's Busiest Day by Order CountSimilar levelHard
Calculate Profit Generated by Each CustomerSimilar levelHard

Same technique, other tables

More practice on this pattern with different columns:

Keep learning

Lesson: SELF JOIN
Lesson: INNER JOIN
Lesson: GROUP BY Multiple Columns
Lesson: Date Functions
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