N031-E3 Tier 3 · Intermediate · easy analytics · Streamhub

Return the user ID and total session count for every user with at least one such session

Part of Chained CTEs in SQL

The problem

Streamhub's product team wants to identify users who have had at least one high-activity session — a session with more than 10 events.

Write a query to return the user ID and total session count for every user with at least one such session.

Assumptions:

  • A user's total session count is the number of sessions records linked to that user_id. A user's peak event count is the largest event_count across those records.
  • A user qualifies when their peak event count is greater than 10.
  • Only qualifying users should appear.

Output:

  • One row per qualifying user, with columns user_id and session_count.
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
WITH
  user_stats AS (
    SELECT
      user_id,
      COUNT(*) AS session_count,
      MAX(event_count) AS max_events
    FROM
      sessions
    GROUP BY
      user_id
  ),
  active_users AS (
    SELECT
      user_id,
      session_count
    FROM
      user_stats
    WHERE
      max_events > 10
  )
SELECT
  user_id,
  session_count
FROM
  active_users

The shape

Two named layers, with the first stage producing two statistics per user and the second stage filtering on one of them while returning the other. The session count and the peak event count are computed side by side in one pass, then the threshold check on max_events decides which users qualify and the session_count is what comes out.

Clause by clause

The first CTE collapses every user's sessions into a single statistics row:

WITH user_stats AS (
  SELECT user_id, COUNT(*) AS session_count, MAX(event_count) AS max_events
  FROM sessions
  GROUP BY user_id
)

GROUP BY user_id produces one row per user. COUNT(*) counts that user's sessions and MAX(event_count) reports the largest event count across those sessions. Both aggregates compute in the same pass, so the cost of carrying the extra column is small.

The second CTE keeps the users whose peak session was a high-activity one:

active_users AS (
  SELECT user_id, session_count
  FROM user_stats
  WHERE max_events > 10
)

WHERE max_events > 10 checks the peak from the first layer. The SELECT list drops max_events because the spec only asks for user_id and session_count; once max_events has done its qualifying work, it does not need to leave the layer.

  • SELECT user_id, session_count FROM active_users returns the qualifying users and their session counts.

The trap

max_events > 10 is a peak check, not an average check. A user with two zero-event sessions and one twelve-event session qualifies because their single high session pushed the MAX above the threshold. Reading the prompt as "average events per session above ten" would lead to AVG(event_count) > 10 and a different result set. The wording matters: "at least one high-activity session" is exactly what MAX(...) > 10 tests.

You practiced layering two WITH stages where the first computes multiple per-user statistics (a count and a max) and the second reads only one of them through a threshold check.

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.