Question

考虑以下数据

test1=# select * from sample order by created_at DESC;
 id | status | service |     created_at
----+--------+---------+---------------------
  8 | OK     |       1 | 2015-09-16 11:54:00
  7 | OK     |       1 | 2015-09-16 11:53:00
  6 | FAIL   |       1 | 2015-09-16 11:52:00
  5 | OK     |       1 | 2015-09-16 11:51:00

如何只选择ID为7和8的行。使用窗口函数我可以获取分区状态的行号，但到目前为止还没有找到如何将结果仅限制为标识为“成功”的最后行的方法给定service的期间。

Answer 1

最简单的方法是：

   SELECT *
   FROM sample AS s
   LEFT JOIN (SELECT service, max(id)
              FROM sample
              WHERE status = 'FAIL'
              GROUP BY service) AS q
   ON s.id > q.id 
      AND s.service = q.service

Answer 2

您需要为每个status = 'FAIL'找到最新service的时间，然后选择最近的service相同的记录：

SELECT *
FROM sample
LEFT JOIN (
  SELECT service, max(created_at) AS last_fail
  FROM sample
  WHERE status = 'FAIL'
  GROUP BY service) f USING (service)
WHERE created_at > last_fail
   OR last_fail IS NULL;     -- also show services without ever failing

这假设只有两个状态代码。如果还有其他内容，请在status = 'OK'子句中添加WHERE过滤器。

仅选择与条件匹配的最后一组行

2 个答案: