N018-M3 Tier 2 · Core SQL · medium analytics · Streamhub

Return the user ID and account plan for every user with no recorded sessions

Part of LEFT JOIN and RIGHT JOIN in SQL

The problem

Streamhub's platform team is auditing user engagement and needs to identify accounts with no session activity.

Write a query to return the user ID and account plan for every user with no recorded sessions.

Assumptions:

  • The users table contains every account on the platform.
  • The sessions table contains every session; user_id links each session back to a user.
  • A user with zero sessions is one whose id does not appear in sessions.user_id.

Output:

  • One row per inactive user, with columns user_id and plan.
Schema · analytics 5 tables
users
id integer
name text
email text
country text
plan text
signed_up_at timestamptz
is_active boolean
conversions
id integer
user_id integer
converted_at timestamptz
plan text
amount numeric
sessions
id integer
user_id integer
started_at timestamptz
ended_at? timestamptz
event_count integer
events
id integer
user_id integer
session_id? integer
event_type text
occurred_at timestamptz
properties? jsonb
periods
id integer
name text
start_month integer
end_month integer

Run previews · Check grades

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

Worked solution Try it yourself first
Solution query
SELECT
  u.id AS user_id,
  u.plan
FROM
  users u
  LEFT JOIN sessions s ON u.id = s.user_id
WHERE
  s.id IS NULL

The shape

The LEFT JOIN keeps every user; the WHERE s.id IS NULL filter then keeps only the users whose session-side columns came back as NULL — meaning the join found no matching session. The 20 rows in the result are exactly the accounts that have never opened a session.

Clause by clause

  • SELECT u.id AS user_id, u.plan returns the user's ID, aliased to user_id, and the user's plan. Both columns are read from the left table, which is intentional: the right table only contributes to the existence check, not to the output.
  • FROM users u LEFT JOIN sessions s ON u.id = s.user_id pairs each user with each of their sessions. Active users produce one row per session with real values in s.*. Inactive users produce a single row with NULL in every s.* column.
  • WHERE s.id IS NULL keeps only the inactive users. sessions.id is the primary key of the sessions table — a real session always has an id. So s.id IS NULL is unambiguous: the row was synthesised by the outer join to preserve an unmatched left-side user.

Why this and not INNER JOIN

INNER JOIN would return only users who do have sessions — the opposite of the question. The audit asks for accounts with no session activity, and INNER JOIN's match-only semantics throws those away before any WHERE clause can pick them out.

The LEFT JOIN is what makes the unmatched users visible in the first place. The WHERE s.id IS NULL then keeps only those visibly-unmatched rows. The two clauses work as a unit: without the join, there's nothing to filter; without the filter, you get every user, matched or not.

The trap

The trap is checking the wrong column for NULL. WHERE u.id IS NULL filters out every row — users.id is a primary key, never NULL on a real row. The query runs cleanly, returns zero rows, and looks like "no inactive users on the platform." The fix is to check a column on the right side of the join — whichever right-side column is guaranteed non-NULL for real rows, with the primary key being the conventional choice.

You practiced the anti-join pattern in a different domain. The structure is identical to 'customers with no orders' — only the table names change.

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.