Question

我有一张桌子

  id  |    name    |         ts
------+------------+---------------------
 3812 | name1      | 2014-05-01 00:10:02
 3900 | name1      | 2014-05-02 00:10:03
 3838 | name2      | 2014-05-01 00:10:08
 3893 | name3      | 2014-05-02 00:10:02
 3933 | name2      | 2014-05-02 00:10:14
 3977 | name3      | 2014-05-03 00:10:01
 3985 | name1      | 2014-05-03 00:10:02
 4006 | name2      | 2014-05-03 00:10:10
 3815 | name3      | 2014-05-01 00:10:02

我需要执行一个只返回最新的选择（通过＆＃39; table.ts＆＃39;），并且只有一个条目用于字段中的每个DISTINCT值＆＃39; table.name＆＃39;

  id  |    name    |         ts
------+------------+---------------------
 3977 | name3      | 2014-05-03 00:10:01
 3985 | name1      | 2014-05-03 00:10:02
 4006 | name2      | 2014-05-03 00:10:10

请帮我完成这项任务。

Answer 1

select id, name, ts
from (
  select id, 
         name, 
         ts, 
         max(ts) over (partition by name) as max_ts
  from the_table
) t
where ts = max_ts;

Answer 2

假设ID按顺序递增...

使用相同名称命名表中的表和列可能会导致以后出现问题...

Select max(n.ID), n.name, max(n.TS) 
From Name n
group by n.name

但是如果ID不是顺序的 - 假设名称和ts是UNIQUE ......

Select a.id, a.name, a.ts
from name a
inner join (Select n.name, max(n.ts) mts from name n group by n.name) b 
  on A.Name = B.Name and A.TS = b.mts

Answer 3

 select b.id, a.name, a.ts 
 from
 (
 select name, max(ts) ts  from table 
 group by name
 ) a
 left join table b on 
 a.id=b.id

PgSQL为列中的每个不同值选择最新的ts

3 个答案: