Hive SQL - 如何从每个人中删除max（date）？

Question

我正在努力编写一种最有效的编写sql查询的方法，即消除每个人的max（day）。我试过where where day < max(day)，但我们的hadoop环境不允许这样做。

基本上，目标是选择在过去570天内切换到同一供应商的具有相同类型电话的人。

对table1查询部分的任何建议？

with table2 as 
(select listener_id, device_id, max(day) day from 
devicetable b
where vendor_id = 42
and category = 'something' 
group by listener_id, device_id, day) -- max day for each person

,table1 as 
(select listener_id, device_id, ROW_NUMBER () over (PARTITION BY listener_id, device_id order by day desc) rowno from
    (select listener_id, device_id, day from devicetable
        where vendor_id=42 and category = 'something'
        group by listener_id, device_id, day)
where rowno <> 1)

insert into finaltable
select a.listener_id
from table1 a
left join 
table2 b
on a.listener_id = b.listener_id
where datediff (a.day, b.day) <=570 and a.day <= b.day -- setting the difference 
and a.device_id <> b.device_id
 and b.listener_id is not null; ```

Answer 1

只是对您的部分代码的第一个建议

如果你想要最大的一天，你不应该提到分组的日期

select listener_id, device_id, max(day) day 
from devicetable b
where vendor_id = 42
and category = 'something' 
group by listener_id, device_id

它是指数据库列还是引用最大值的别名

Answer 2

尝试这样的事情：

SELECT listener_id, device_id, day FROM devicetable AS A 
LEFT OUTER JOIN (SELECT listener_id, Max(Day) AS MaxDate FROM devicetable GROUP BY listener_id) AS B ON A.listener_id=B.listener_id AND A.day=B.MaxDate
WHERE B.listener_id IS NULL

它将返回除每个侦听器的最大日期之外的所有行。