Question

您好以下案例有问题。我有一张表＆＃34;应用程序日志＆＃34;如

App_name        date         status
Application 1   10-MAR-17   SUCCEEDED
Application 1   11-MAR-17   SUCCEEDED
Application 1   12-MAR-17   FAILED
Application 1   13-MAR-17   SUCCEEDED
Application 1   14-MAR-17   SUCCEEDED
Application 1   15-MAR-17   FAILED
Application 1   16-MAR-17   SUCCEEDED
Application 1   17-MAR-17   SUCCEEDED
Application 1   18-MAR-17   FAILED
Application 1   19-MAR-17   SUCCEEDED
Application 1   20-MAR-17   SUCCEEDED
Application 1   21-MAR-17   FAILED
Application 1   22-MAR-17   SUCCEEDED
Application 1   23-MAR-17   SUCCEEDED
Application 1   25-MAR-17   SUCCEEDED
Application 3   20-MAR-17   FAILED
Application 3   21-MAR-17   FAILED
Application 3   22-MAR-17   FAILED
Application 3   23-MAR-17   FAILED
Application 3   24-MAR-17   FAILED

我必须找到＆＃39; n＆＃39;一个application_name的后续失败状态并将其作为结果返回。例如，如果n = 5，我的查询必须返回app_name = Application 3。

Answer 1

以下查询适用于SQL Server。我也没有看到为什么它也不应该在Oracle中工作的原因：

SELECT [App_name]
FROM (
   SELECT [App_name], [date], [status],
          ROW_NUMBER() OVER (PARTITION BY App_name ORDER BY [date]) 
          -
          ROW_NUMBER() OVER (PARTITION BY App_name, 
                                          CASE 
                                             WHEN status = 'FAILED' THEN 1
                                             ELSE 2 
                                          END 
                             ORDER BY [date]) AS grp
   FROM ApplicationLog) AS t
WHERE status = 'FAILED'   
GROUP BY [App_name], grp
HAVING COUNT(*) >= 5

Demo here

Answer 2

这是另一种仅使用单个OLAP函数的方法：

SELECT App_name, Max(date) AS LastFail
FROM
 (
   SELECT App_name, date, status,
       -- check if the last 5 rows only contain FAILED
          Min(CASE WHEN status = 'FAILED' THEN 1 ELSE 0 end) 
          Over (PARTITION BY App_name
                ORDER BY date
       -- caution, as this includes the current row it must be n-1
                ROWS 4 Preceding) AS x
   FROM ApplicationLog
 ) AS t
WHERE x = 1
GROUP BY App_name

一列中具有相同值的多行

2 个答案: