HardSQL challenge #205WINDOW FUNCTIONLAGDATE

Calculate Days Between Consecutive Orders per Customer with LAG()

Return every order with the number of days since the same user's previous 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.

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The task

Find days between consecutive orders per user.

A correct answer returns 200 rows with 3 columns: user_id, order_date, gap.

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_idorder_dategap
12022-09-26NULL
22023-04-02NULL
42022-08-22NULL
42024-04-20607
42024-11-15209

First 5 of 200 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

  • 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

LAG()

Show the step-by-step approach

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

  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 gap: julianday(order_date) - julianday(LAG(order_date) OVER (PARTITION BY user_id ORDER BY order_date)).

Model solution

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What to watch for

LAG() reads the previous row inside the same partition. Each customer's first order has no previous row, so its gap is NULL rather than 0. Subtracting two julianday() values gives the difference in days.

  • 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.
  • Date functions. These date functions are SQLite syntax. MySQL uses DATE_FORMAT and DATEDIFF, PostgreSQL uses TO_CHAR, EXTRACT and date subtraction.
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Same technique, other tables

More practice on this pattern with different columns:

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