Top Users by Orders
Find users who placed the most orders.
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 users who placed the most orders.
A correct answer returns 114 rows with 2 columns: user_id, total_orders.
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 | total_orders |
|---|---|
First 5 of 114 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 rows with GROUP BY to get one result row per group
- Sorting the result with ORDER BY
Hint and approach
Try the challenge first. Open these only when you need them.
Show the hint
GROUP BY ORDER BY DESC
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
COUNT(*). - Sort the result by
total_ordersdescending.
Model solution
Checking your session…
What to watch for
- 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.
- ORDER BY. Without ORDER BY the row order is not guaranteed, even if it looks stable. Rows that tie on the sort columns can come back in any order.
- 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
- Salary Statistics per Department: Count, Avg, Min and MaxSimilar levelMedium
- Find the Best-Selling Products by QuantitySimilar levelMedium
- Count Orders by Status with GROUP BYEasierEasy
- Rank Products by Their Average Review RatingHarderHard
- Calculate Monthly Revenue with strftimeSimilar levelMedium
Keep learning
- Lesson: GROUP BY
- Lesson: ORDER 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