EasySQL challenge #102GROUP BY

Payments by Method

Count payments grouped by method.

The challenge runs on the SQLabHub practice database (SQLite) and uses the payments table. You can solve it in the browser, with nothing to install.

Solve This SQL Challenge

The task

Count payments grouped by method.

A correct answer returns 8 rows with 2 columns: method, COUNT(*).

Tables used in this challenge

`payments` 200 rows

id INTEGER (PK) · order_id INTEGER · user_id INTEGER · amount REAL · method TEXT · status TEXT · paid_at TEXT · transaction_id TEXT

First 3 rows of the payments table
id	order_id	user_id	amount	method	status	paid_at	transaction_id
1	46	105	921.62	Gift Card	success	2022-05-18	TXNHTN4GIK5L
2	157	37	1723.82	Bank Transfer	success	2024-11-11	TXNKJPIXJOFP
3	45	36	978.76	Bank Transfer	refunded	2023-10-24	TXNU2RW4MK1R

The first rows of each table are shown. See how all the tables relate on the database schema page.

Expected output

Expected output of the challenge
method	COUNT(*)
Apple Pay	27
Bank Transfer	38
Credit Card	14
Cryptocurrency	34
Debit Card	28
Gift Card	22
Google Pay	13
PayPal	24

The complete result: 8 rows. This is the output of the model solution on the practice database. The task does not fix a sort order, so your rows may come back in a different order.

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

Hint and approach

Try the challenge first. Open these only when you need them.

Show the hint

GROUP BY method

Show the step-by-step approach

How the model solution is built, in the order the database evaluates it:

Read from the payments table.
Group the rows by method, so the result has one row for each value.
For each group, compute COUNT(*).

Model solution

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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.
Aggregate functions. Aggregates skip NULLs. AVG divides by the number of non-NULL values and COUNT(column) can be smaller than COUNT(*).