HardSQL challenge #353JOINDATEGROUP BY

Count Orders in Each Customer's First 7 Days

Return each user with the number of orders placed within seven days of their first order.

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 most orders within first 7 days of first order.

A correct answer returns 114 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
124	2
47	2
150	1
149	1
148	1

First 5 of 114 rows. This is the output of the model solution on the practice database.

What you practise

Filtering rows with a WHERE condition
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
Combining rows from two tables on a matching key with an inner join
Querying the result of another query with a subquery in FROM
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

MIN(order_date) + 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.
Build a second inner query (f) to join to:
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 MIN(order_date).
Join that inner query on orders.user_id = f.user_id.
Filter in WHERE: keep the rows where orders.order_date is at most 7 days after f.first_date.
Group the rows by orders.user_id, so the result has one row for each value.
For each group, compute COUNT(*).
Sort the result by cnt descending.

Model solution

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What to watch for

The derived table finds every customer's first order date. Joining back to orders and filtering with DATE(first_date, '+7 day') keeps only the early orders. The first order itself is always counted, so the minimum is 1.

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