HardSQL challenge #283WINDOW FUNCTIONLAGDATE

Find Customers Who Never Went More Than 7 Days Between Orders

Return the users whose longest gap between consecutive orders is seven days or less.

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 who placed orders at least once every 7 days.

A correct answer returns 1 row with one column: user_id.

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
iduser_idtotaldiscountitem_countstatusorder_dateshipped_dateshipping_cityshipping_country
1432638.110.253cancelled2024-04-24NULLLos AngelesUSA
2829470.195delivered2024-03-152023-04-09ChicagoUSA
3136683.240.082delivered2023-02-152023-07-10Los AngelesUSA

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
47

The complete result: 1 row. This is the output of the model solution on the practice database.

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
  • Querying the result of another query with a subquery in FROM
  • Working with dates in SQLite: strftime, julianday and DATE
  • Reading the previous row's value with LAG()
  • Restarting a window calculation for each group with PARTITION BY
  • Deriving a new column from an expression and naming it with an alias

Hint and approach

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

Show the hint

julianday difference

Show the step-by-step approach

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

  1. Build an inner query first (a derived table):
    1. Read from the orders table.
    2. Fetch order_date from the previous row with LAG(), ordered by order_date, restarting for each user_id. Where there is no such row the value is NULL.
    3. Combine those values into diff: julianday(order_date) - julianday(LAG(order_date) OVER (PARTITION BY user_id ORDER BY order_date)).
  2. Group the rows by user_id, so the result has one row for each value.
  3. For each group, compute MAX(diff).
  4. Filter the groups in HAVING: keep those where MAX(diff) is at most 7.

Model solution

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What to watch for

Saying every gap is at most 7 is the same as saying the largest gap is at most 7, which is what HAVING MAX(diff) <= 7 tests. Users with only one order have no gaps, so their maximum is NULL and they are left out.

  • LAG(). LAG() returns NULL for the first row of each partition, and any arithmetic with that NULL is NULL too. Supply a default as the third argument if you need a number.
  • 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.
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