Find the Last Order Date for Each Customer
Return each user with the date of their most recent 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.
The task
Find last order date for each user.
A correct answer returns 114 rows with 2 columns: user_id, MAX(order_date).
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
| id | user_id | total | discount | item_count | status | order_date | shipped_date | shipping_city | shipping_country |
|---|---|---|---|---|---|---|---|---|---|
The first rows of each table are shown. See how all the tables relate on the database schema page.
Expected output
| user_id | MAX(order_date) |
|---|---|
First 5 of 114 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
Hint and approach
Try the challenge first. Open these only when you need them.
Show the hint
MAX(order_date)
Show the step-by-step approach
How the model solution is built, in the order the database evaluates it:
- Read from the
orderstable. - Group the rows by
user_id, so the result has one row for each value. - For each group, compute
MAX(order_date).
Model solution
Checking your session…
What to watch for
MAX works on dates because they are stored as sortable YYYY-MM-DD text. This returns the date only; getting the rest of that order's row needs a join back, a correlated subquery or ROW_NUMBER().
- GROUP BY. Every selected column must either be in GROUP BY or sit inside an aggregate. SQLite accepts other columns and returns an arbitrary value for them; most databases raise an error.
- Aggregate functions. Aggregates skip NULLs. AVG divides by the number of non-NULL values and COUNT(column) can be smaller than COUNT(*).
Related SQL challenges
- Summarise Each Order: Total Quantity and AmountSimilar levelMedium
- Count Orders per Month with strftimeSimilar levelMedium
- Total Sales per Order Status with SUM and GROUP BYEasierEasy
- Calculate Each Customer's Order Range (Max Minus Min)HarderHard
- Count Employees in Each DepartmentEasierEasy
Keep learning
- Lesson: GROUP BY
- Lesson: Aggregate Functions
- All SQL practice challenges, by difficulty and topic
- SQL interview questions with answers
- SQL Playground: run any query on the same database