SQLize Online / PHPize Online  /  SQLtest Online

A A A
Share      Blog   Popular
Copy Format Clear
create table t (client_id int, time_key date); insert into t values (1, '2022-01-01'), (2, '2022-02-01'), (22, '2022-02-01'), (3, '2022-03-01'), (3, '2022-03-01'), (4, '2022-04-01'), (5, '2022-05-01'), (6, '2022-06-01'), (7, '2022-07-01'), (77, '2022-07-01'), (8, '2022-08-01'), (9, '2022-09-01'), (10, '2022-10-01'), (11, '2022-11-01'), (12, '2022-12-01'); select date_part('year', time_key) as "year", date_part('month', time_key) as "month", sum(cnt) over(order by time_key rows between 2 preceding and current row) as cumcount from ( select time_key, count(distinct client_id) as cnt from t group by 1 )foo
Stuck with a problem? Got Error? Ask ChatGPT!
Copy Clear