在Impala中按组减去最大，最小日期

Posted 2021-03-26

我有一个查询，我在impala中找到用户的最大和最小日期：

select max(id_date) as last_tran
       ,min(id_date) as first_tran
       ,user
from table_1 a 
join table_2 d
on a.id = d.id
group by 3

我想再按用户减去最小和最大日期。

在Impala中我尝试使用date_sub函数，但它不起作用。

select date_sub(last_tran, first_tran) as date_len
, user
from
(select max(id_date) as last_tran
       ,min(id_date) as first_tran
       ,user
from table_1 a 
join table_2 d
on a.id= d.id
group by 3) time
group by 1,2

似乎date_sub函数的第二个参数必须是一个表示天数的整数。

我怎么能绕过这个？

答案

为了使这项工作，datediff函数应该像这样使用：

select datediff(last_tran, first_tran) as date_len
, user
from
(select max(id_date) as last_tran
       ,min(id_date) as first_tran
       ,user
from table_1 a 
join table_2 d
on a.id= d.id
group by 3) time
group by 1,2

该函数返回开始日期和结束日期之间的天数。从documentation看来，天的回报值似乎无法改变。