HardSQL challenge #241WINDOW FUNCTIONLAGPATTERN

Find Customers Whose Every Order Is Larger Than the Last

Return the users with at least two orders whose order total went up every time.

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 with at least two orders whose every order total is higher than the one before it.

A correct answer returns 20 rows 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
10
22
32
43
44

First 5 of 20 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

  • 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
  • Reading the previous row's value with LAG()
  • 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

Compare each order with LAG(total), then require that every comparison per user is an increase

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 total 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.
  2. Group the rows by user_id, so the result has one row for each value.
  3. For each group, compute COUNT(prev) and SUM(total > prev).
  4. Filter the groups in HAVING: keep those where COUNT(prev) is greater than 0, and SUM(total > prev) equals COUNT(prev).

Model solution

Checking your session…

What to watch for

Requiring every step to be an increase means the number of increases must equal the number of comparisons. COUNT(prev) counts the comparisons (the first order has no previous one), and SUM(total > prev) counts the increases, because a comparison is 1 or 0 in SQLite. Without the COUNT(prev) > 0 test, single-order users would pass with zero of zero.

  • 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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Same technique, other tables

More practice on this pattern with different columns:

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