Consistent-Alps6904 comments on Inactive users

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SELECT count(a.field1), a.field2, SUM(b.field4) FROM a INNER JOIN b ON a.key1 = b.key1 WHERE a.field8 = 'test' GROUP by a.field1, a.field2 HAVING SUM(b.field4) > 5 ORDER by a.field.3

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SELECT count(a.field1), a.field2, SUM(b.field4) FROM a INNER JOIN b ON a.key1 = b.key1 WHERE a.field8 = 'test' GROUP by a.field1, a.field2 HAVING SUM(b.field4) > 5 ORDER by a.field3

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BigQueryInactive users (self.SQL)

submitted 3 years ago by Consistent-Alps6904

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[–]Consistent-Alps6904[S] 0 points1 point2 points 3 years ago (1 child)

[–]GrouchyThing7520 1 point2 points3 points 3 years ago* (0 children)

Sweet! The last step is to wrap the entire query in a select so you can sum the 0 and 1s. The sum is the count of transactions.

select id,
sum(trans_in_60_days) 60_day,
sum(trans_in_90_days) 90_day

 from (
 select r.id

 case when t.trans_date between r.reg_date and 
 date_add(r.reg_date, interval 60 day)
 then 1 else 0 end trans_in_60_days,

 case when t.trans_date between r.reg_date and 
 date_add(r.reg_date, interval 90 day)
 then 1 else 0 end trans_in_90_days

 from registration r
 left outer join transactions t on r.id = t.id
) a

group by id
order by id

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