Question

SELECT id, server_id, start_time, end_time 
FROM errors 
WHERE server_id in (3, 12, 24, 25, 26, 27, 28, 29, 30) 
ORDER BY id DESC 
LIMIT 9

这是我试图运行的查询，以便为server_id = 3, 12, 24, 25, 26, 27, 28, 29, 30提供结果。相反，我收到的是server_id = 25, 25, 12, 25, 27, 27, 28, 28, 27。请注意重复的server_ids。该查询为我提供了唯一的id但重复server_id。

我是否有办法获得可以显示每个id的最后server_id的结果？

我尝试过做ORDER BY server_id，但这给了我同样的问题。

我试过运行DISTINCT，但这也行不通。

Answer 1

您遇到的问题是每个服务器只需要一条记录，其中包含最大ID ..和相关信息。您需要将结果限制为最大ID ...这是单向...

SELECT id, server_id, start_time, end_time 
FROM errors 
WHERE server_id in (3, 12, 24, 25, 26, 27, 28, 29, 30) 
and ID = (Select max(ID) from errors E2 where E2.server_ID=errors.server_ID)
ORDER BY id DESC 
LIMIT 9

Answer 2

你必须使用一些聚合函数。

像

这样的东西

select 
  server_id,
  max(id),
  avg(start_time),--for example
  avg(end_time)--for example
from errors
where server_id in (3, 12, 24, 25, 26, 27, 28, 29, 30) 
group by server_id
order by id desc

如果你需要server_id对应于max id的start_time和end_time，你可以这样做

select e.id, e.server_id, e.start_time, e.end_time
from errors e
join (select server_id, max(id) maxid
      from errors
      group by server_id) t
  on t.maxid = e.id and e.server_id = t.server_id
where e.server_id in (3, 12, 24, 25, 26, 27, 28, 29, 30) 
order by e.id DESC

MySQL在哪里IN和区别和限制

2 个答案: