Question

我有一个查询，它给了我以下结果。此结果集包含服务器上的作业启动时以及完成时的时间。

JobName             LatestStartTime         LatestEndTime
Job1                2016-04-15 00:00:40.000 2016-04-15 00:07:40.000
Job2                2016-04-15 00:01:23.000 2016-04-15 00:17:37.000
Job3                2016-04-15 08:00:03.000 2016-04-15 08:18:05.000
Job4                2016-04-15 08:30:06.000 2016-04-15 08:57:21.000
Job5                2016-04-15 09:00:03.000 2016-04-15 09:07:49.000
Job6                2016-04-15 03:53:40.000 2016-04-15 03:53:41.000
Job7                2016-04-15 09:30:07.000 2016-04-15 11:36:35.000

另一方面，我有一个查询，创建一个间隔15分钟的临时表。如下：

Increment
2016-04-15 00:00:00.000
2016-04-15 00:15:00.000
2016-04-15 00:30:00.000
2016-04-15 00:45:00.000
2016-04-15 01:00:00.000
2016-04-15 01:15:00.000
2016-04-15 01:30:00.000
2016-04-15 01:45:00.000
2016-04-15 02:00:00.000
2016-04-15 02:15:00.000
2016-04-15 02:30:00.000
2016-04-15 02:45:00.000

我想知道哪些工作以确定的增量运行。例如，我的最终结果集应该类似于：

Increment                     NumberOfJobs    JobNames
2016-04-15 00:00:00.000       2               Job1, Job2
2016-04-15 00:15:00.000       1               Job2
2016-04-15 00:30:00.000       0               NULL

OR

Increment                     NumberOfJobs    JobNames
2016-04-15 00:00:00.000       2               Job1
2016-04-15 00:00:00.000       2               Job2

2016-04-15 00:15:00.000       1               Job2
2016-04-15 00:30:00.000       0               NULL

Answer 1

困难的部分是将工作列表放在一行上。计数很简单：

select t.increment, j.jobname,
       sum(count(*)) over (partition by t.increment) as countOfJobs
from times t left join
     jobs j
     on t.increment >= j.lasteststarttime and
        t.increment <= j.lastestendtime
group by t.increment, j.jobname
order by increment, jobname;

获取列表需要在SQL Server中使用奇怪的子查询：

select t.increment, count(*) as numJobs,
       stuff((select ', ' + j2.jobname
              from jobs j2
              where t.increment >= j2.lasteststarttime and
                    t.increment <= j2.lastestendtime
              for xml path ('')
             ), 1, 1, '') as jobs
from times t  left join
         jobs j
         on t.increment >= j.lasteststarttime and
            t.increment <= j.lastestendtime
group by t.increment
order by increment;

如果作业名称包含异常字符（那些需要转义为XML的字符），则XML逻辑稍微复杂一些。但是，对于职位名称来说，这似乎不太可能。

Answer 2

试试这个：

SELECT [increment], JobName
, Count(1) over (Partition by [increment]) AS NumberOfJobs    
FROM temp_table
INNER JOIN (
  SELECT JobName, st.t AS t1, LatestEndTime AS t2
  FROM query AS q
  OUTER APPLY (
    SELECT Max([increment]) AS t 
    FROM temp_table 
    WHERE [increment] < q.LatestStartTime) as st
  ) as j ON Increment BETWEEN j.t1 and j.t2

它返回：

increment   JobName NumberOfJobs
2016-04-15 00:00:00.000 Job1    2
2016-04-15 00:00:00.000 Job2    2
2016-04-15 00:15:00.000 Job2    1
2016-04-15 02:45:00.000 Job3    5
2016-04-15 02:45:00.000 Job4    5
2016-04-15 02:45:00.000 Job5    5
2016-04-15 02:45:00.000 Job6    5
2016-04-15 02:45:00.000 Job7    5

左连接：

SELECT [increment], JobName
, Count(JobName) over (Partition by [increment]) AS NumberOfJobs    
FROM temp_table
LEFT JOIN (
  SELECT JobName, st.t AS t1, LatestEndTime AS t2
  FROM query AS q
  OUTER APPLY (
    SELECT Max([increment]) AS t 
    FROM temp_table 
    WHERE [increment] < q.LatestStartTime) as st
  ) as j ON Increment BETWEEN j.t1 and j.t2

返回：

2016-04-15 00:00:00.000 Job1    2
2016-04-15 00:00:00.000 Job2    2
2016-04-15 00:15:00.000 Job2    1
2016-04-15 00:30:00.000 NULL    0
2016-04-15 00:45:00.000 NULL    0
2016-04-15 01:00:00.000 NULL    0
2016-04-15 01:15:00.000 NULL    0
2016-04-15 01:30:00.000 NULL    0
2016-04-15 01:45:00.000 NULL    0
2016-04-15 02:00:00.000 NULL    0
2016-04-15 02:15:00.000 NULL    0
2016-04-15 02:30:00.000 NULL    0
2016-04-15 02:45:00.000 Job3    5
2016-04-15 02:45:00.000 Job4    5
2016-04-15 02:45:00.000 Job5    5
2016-04-15 02:45:00.000 Job6    5
2016-04-15 02:45:00.000 Job7    5

Answer 3

SELECT  *,
        COUNT(*) OVER (PARTITION BY increment) numberOfJobs
FROM    times
JOIN    jobs
ON      increment BETWEEN latestStartTime AND latestEndTime

根据时间戳进行计数

3 个答案: