N047-M2 Tier 4 · Advanced · medium ecommerce · Brightlane

Return the ID and name of every customer who has placed more than `3` orders on record

Part of Correlated Subqueries in SQL

The problem

Brightlane's loyalty program team is identifying high-engagement customers for a rewards campaign.

Write a query to return the ID and name of every customer who has placed more than 3 orders on record.

Assumptions:

  • A customer's order count is the number of orders linked to that customer_id.
  • Only customers whose order count is strictly greater than 3 should appear.

Output:

  • One row per qualifying customer, with columns id and name.
Schema · ecommerce 5 tables
categories
id integer
name text
parent_id? integer
products
id integer
name text
category_id integer
price numeric
stock_qty integer
attributes? jsonb
order_items
id integer
order_id integer
product_id integer
quantity integer
unit_price numeric
customers
id integer
name text
email text
city? text
country text
created_at timestamptz
is_active boolean
orders
id integer
customer_id integer
ordered_at timestamptz
status text
total_amount numeric

Run previews · Check grades

Write a query, then run it to see results here.

Worked solution Try it yourself first
Solution query
SELECT
  c.id,
  c.name
FROM
  customers c
WHERE
  (
    SELECT
      COUNT(*)
    FROM
      orders o
    WHERE
      o.customer_id = c.id
  ) > 3

The shape

The qualifying condition is "this customer has more than three orders," which is a per-customer count compared to a constant. A correlated \COUNT(*)\ in the \WHERE\ clause computes that count once per customer and the outer query keeps only the customers where it exceeds \3\.

Clause by clause

  • \SELECT c.id, c.name\ returns the two columns the spec asks for, taken from the customer row.
  • \FROM customers c\ reads every customer. The outer alias \c\ is the row the inner subquery correlates against.
  • \WHERE (SELECT COUNT(*) FROM orders o WHERE o.customer_id = c.id) > 3\ is the correlated filter. The inner \WHERE o.customer_id = c.id\ ties the count to the outer customer, so the subquery returns one integer per customer. The outer comparison \> 3\ keeps the customer only when that integer is strictly greater than three. A customer with exactly four orders qualifies; a customer with exactly three does not, which matches the "strictly greater than \3\" wording.

Why this and not a \JOIN\ plus \GROUP BY\ plus \HAVING\

\SELECT c.id, c.name FROM customers c JOIN orders o ON o.customer_id = c.id GROUP BY c.id, c.name HAVING COUNT(*) > 3\ returns the same customers. The two forms are interchangeable for this question, and the planner often rewrites one into the other internally. The correlated form is shorter and does not require a \GROUP BY\, which is sometimes the more readable expression when the answer is a yes/no filter on a per-outer-row count rather than a grouped aggregation that needs to be displayed.

You practiced a correlated COUNT(*) in WHERE — the per-customer count is computed inside the predicate, so the outer record is kept only when the count beats the threshold.

How you actually get good at SQL

Reading explains SQL. Writing it, over and over with instant feedback, is what makes you fluent.

That's the whole SQLMaxx loop: 600+ real problems, instant AI feedback, mastery you can actually see, and spaced review that won't let you forget.

A stack of SQL practice problem cards, the top card showing an employees table.
615 problems · 66 concepts

Real problems. Not toy examples.

615 hand-built problems spanning all 66 concepts, from basic SELECTs to window functions, built on real schemas and real business questions, the kind you'll actually get asked on the job. Enough reps to make SQL automatic.

A retro computer showing a SQL query marked correct with a green checkmark.
Instant AI feedback

Write a query. Know if it's right in one second.

No copying an answer and hoping it clicked. The AI grader checks your real query against real data, catches exactly what's wrong, and explains the fix in plain English, like a senior analyst reading over your shoulder on every problem.

A circular mastery progress dial filling from blue to green, the SQLMaxx diamond at its center.
Mastery tracking

Stop guessing whether you actually know it.

SQLMaxx tracks every concept and shows you what you've mastered and what's still shaky. Your skills fill in one concept at a time, so 'I think I get joins' becomes something you can prove.

A SQL query editor circled by a blue return arrow with a clock, scheduled to come back for review.
Spaced review

Learn it once. Keep it for good.

Most of what you learn this week fades by next week. So when a concept comes due for review, SQLMaxx hands you a fresh problem to solve from a blank editor, not a flashcard to re-read. A research-backed spaced-repetition algorithm (FSRS) times each return for right before you'd forget, so your SQL is still there months later, when the interview or the job actually needs it.

Practice, feedback, mastery, review. That's the loop that turns reading into real skill.

Start free

No account, no credit card. Start solving in under a minute.