HardSQL challenge #291WINDOW FUNCTIONPATTERNGROUP BY

Find the Longest Run of Consecutive Order Days per Customer

Return the length of each run of back-to-back order days for every user, longest 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.

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The task

Find users with longest consecutive daily ordering streak.

A correct answer returns 199 rows with 2 columns: user_id, streak.

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_idstreak
472
11
21
41
41

First 5 of 199 rows. 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 by more than one column to get a row per combination
  • Querying the result of another query with a subquery in FROM
  • Working with dates in SQLite: strftime, julianday and DATE
  • Numbering rows with ROW_NUMBER() to pick one or a few per group
  • Restarting a window calculation for each group with PARTITION BY
  • Sorting the result with ORDER 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

date difference grouping trick

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. Number the rows with ROW_NUMBER(), ordered by order_date, restarting for each user_id.
    3. Combine those values into grp: julianday(order_date) - ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY order_date).
  2. Group the rows by user_id and grp, so the result has one row for each combination.
  3. For each group, compute COUNT(*).
  4. Sort the result by streak descending.

Model solution

Checking your session…

What to watch for

Within a run of consecutive days, the day number minus the row number is constant, so that difference identifies the run. Grouping by it and counting gives the run length. Two orders on the same day would split a run; this dataset has none.

  • 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.
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Same technique, other tables

More practice on this pattern with different columns:

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