Question

我正在努力写一点写SQL查询的最有效方法，这消除了每个人的max（day）。我尝试了where where day < max(day)，但我们的hadoop环境不允许这样做。

基本上，目标是选择一个在过去570天内切换到同一供应商的手机类型相同的人。

对查询的table1部分有任何建议吗？

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

仅是部分代码的第一个建议

如果您想要最多的日期，则不要在分组依据中提及日期

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

是引用数据库列还是引用max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

它将返回所有行，但不包括每个侦听器具有最大日期的行。

Hive SQL-如何从每个人中删除最大值（日期）？

2 个答案: