HardSQL challenge #218WINDOW FUNCTIONLAGPATTERN

Find Customers Who Ordered on Consecutive Days

Return the users who placed orders on two days in a row.

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 consecutive order streak.

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

  • Choosing exactly which columns a query returns
  • Filtering rows with a WHERE condition
  • 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

LAG date diff

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. Filter in WHERE: keep the rows where diff equals 1. The filter sits in the outer query because a window function cannot be tested in the WHERE clause of the query that computes it.
  3. Return only the column user_id.

Model solution

Checking your session…

What to watch for

LAG() puts each order next to the same customer's previous one, and a difference of exactly 1 between the julianday values means back-to-back days. A user appears once for every such pair unless you add DISTINCT.

  • 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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