Independent practice · SQLite

Identify visitors without checkout

Your task

Return customer_id for visitors who never checked out.
No repeated customer IDs.

Target: 70/100. Suggested active time: 10 minutes. You can keep working after that.

Show a hint

Use NOT EXISTS or a suitable left join.

View the scoring standard

Accuracy checks the values. Coverage checks the requested output. Time is recorded separately and never changes your score.

columns: 20 points

rows: 80 points

Tables and practice data

orders

order_id	customer_id	status	amount	region	ordered_at
1	1	paid	120	East	2026-09-01
2	2	pending	140	West	2026-09-02
3	1	paid	100	East	2026-09-03
4	3	paid	180	West	2026-09-03
5	99	paid	80	East	2026-09-04
6	NULL	cancelled	0	NULL	2026-09-04
7	2	paid	120	West	2026-09-05
8	4	paid	30	North	2026-09-06

customers

customer_id	name	email
1	Avery	team@example.test
2	Morgan	team@example.test
3	Riley	NULL
4	Casey	(empty)
5	Sky	sky@example.test

returns

return_id	order_id	refund
1	1	20
2	1	10
3	4	180
4	99	25

inventory

sku	on_hand	reorder_at
PEN	5	10
BOOK	20	10
MUG	0	5
BAG	NULL	4
CARD	4	4

events

event_id	customer_id	stage
1	1	visit
2	1	visit
3	1	checkout
4	2	visit
5	3	visit
6	3	checkout
7	4	visit

Download the input →

Your work

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SQL query

Import answer

Show the reference answer

SELECT DISTINCT e.customer_id FROM events e WHERE e.stage='visit' AND NOT EXISTS(SELECT 1 FROM events c WHERE c.customer_id=e.customer_id AND c.stage='checkout')

Customers 2 and 4 visited but have no checkout event.