postgres / sql 自连接比较行

Question

我有以下示例表

id, label, reason, timestamp
7, 123, ok, 10:50:01
6, 124, reject, 10:53:01
5, 123, reject, 10:30:02
4, 124, ok, 10:30:01

我想获取如果标签的最后一条记录（最新时间戳）是可以的行，我们会在结果中显示它，但如果有拒绝，我们不会。

例如

7, 123, ok, 10:50:01

这应该很简单，但我真的有问题。 有人可以帮忙吗？ 提前致谢。

Answer 1

我建议在以下方面distinct on ：

select t.*
from (select distinct on (label) t.*
      from t
      order by label, timestamp desc
     ) t
where reason = 'ok';

你也可以这样做：

select t.*
from t
where t.timestamp = (select max(t2.timestamp)
                     from t t2
                     where t2.label = t.label
                    ) and
      t.reason = 'ok';

Answer 2

这是您的查询。

select t1.* from 
  (select row_number() over (order by timestamp desc) as rn
       , id, label, reason, timestamp
   from tableA) as t1 
where t1.rn = 1 and t1.reason = 'ok'

Answer 3

我们可以在这里使用一些存在的逻辑：

SELECT id, label, reason, timestamp
FROM yourTable t1
WHERE
    reason = 'ok' AND
    NOT EXISTS (SELECT 1 FROM yourTable t2
                WHERE t2.label = t1.label AND
                      t2.timestamp > t1.timestamp);

这假设您想要具有最新时间戳的匹配单行。