我正在做很长的查询,以查找在某些日期具有特定条件的客户,在这种情况下,'2019-6-20'
,查询就是这样
这是我的代码
select current_date() as date , count(customer_id) as cell13
from(
select customer_id, count(id) as total, string_agg(payment_state order by created_at desc limit 1) as cek
from(
select distinct(A.id), A.customer_id, extract(month from A.created_at) as months,extract(day from A.created_at) as days, extract(year from A.created_at) as years, payment_state, A.created_at, A.grandtotal_cents
from bl.orders as A
left join bl.blacklists as B
on A.customer_id = B.customer_id
where date(A.created_at) >= date_sub(date('2019-6-20') , interval 60 day) and grandtotal_cents > 0 and B.customer_id is null
)
group by customer_id
having cek = "unpaid")
这是结果
Row date cell13
1 2019-06-21 696
现在,我需要查询特定日期范围内的多个日期,例如2019-03-23
至2019-06-21
。我该怎么做,所以输出会喜欢
Row date cell13
1 2019-06-21 696
...
90 2019-03-23 ...
答案 0 :(得分:1)
您可以使用generate_date_array()
和unnest()
生成日期表,然后将其与left join
一起使用。
总的来说,您的查询是一条难以理解的消息,但这是个主意:
with dates as (
select dte
from (select generate_date_array('2019-03-23', '2016-06-21', interval 1 day) d
) d cross join
unnest(d.d) dte
)
select . . .
from dates left join
bl.orders o
on date(o.created_at) >= date_sub(dte, interval 60 day)
. . .