从 group by 中获取一条记录，而无需手动键入所有列

Question

我有一个 table 以下列：

id, col_1, col_2, col_3, .... col_1000, timestamp

我只想为每个 id 记录一条记录，并采用最新的时间戳。也就是说，我的查询是：

select id, max_by(col_1, timestamp), max_by(col_2, timestamp), max_by(col_3, timestamp), ... max_by(col_1000, timestamp), max_by(timestamp, timestamp) group by id

因为有这么多专栏，我想知道有没有办法可以做这样的事情：

select max_by(*, timestamp) group by id

这样我就不必手动输入所有列了？谢谢！

Answer 1

嗯。 . .您可以使用 row_number():

select t.*
from (select t.*,
             row_number() over (partition by id order by timestamp desc) as seqnum
      from t
     ) t
where seqnum = 1;

您也可以使用相关子查询或 join:

来表达这一点

select t.*
from t join
     (select id, max(timestamp) as max_timestamp
      from t
      group by id
     ) tt
     on t.id = tt.id and t.timestamp = tt.max_timestamp

从 group by 中获取一条记录，而无需手动键入所有列

Taking one record from group by without manually typing all columns

sql

hiveql

presto