Question

您好我想知道基于下表的作业执行时间。例如，1/24/17，“运行”和“已完成”之间的时差为17分钟。

**JobName**     **CurrentJobStatus**    **TimeOfChange**
    Customer    Completed           1/24/17 0:17
    Customer    Running             1/24/17 0:00
    Customer    Waiting             1/23/17 20:30
    Customer    Completed           1/21/17 0:47
    Customer    Idle               1/21/17 16:00
    Customer    Running             1/21/17 0:30
    Customer    Completed           1/20/17 0:31
    Customer    Idle                1/20/17 16:00
    Customer    Running             1/20/17 0:15
    Customer    Waiting             1/20/17 20:30
    Customer    Completed           1/19/17 0:37
    Customer    Idle                1/19/17 16:00
    Customer    Running             1/19/17 0:20
    Customer    Waiting             1/19/17 20:30

Answer 1

假设每天只有1 Running和Completed个工作状态，并且您想要两者之间的时差，您可以通过添加case语句来使用条件聚合在max或min内：

select * , datediff(minute, completed, running) 
from (
    select 
        convert(date, TimeOfChange) change_date,
        max(case when CurrentJobStatus = 'Completed' then TimeOfChange end) completed,
        min(case when CurrentJobStatus = 'Running' then TimeOfChange end) running
    from mytable
    group by convert(date, TimeOfChange)
) t order by change_date

Answer 2

我已选择使用CurrentJobStatus的行（＆＃39;已完成＆＃39;，＆＃39;正在运行＆＃39;），然后我按TimeOfChange排序的CurrentJobStatus设置排名。这为每对（＆＃39;已完成＆＃39;，＆＃39;正在运行＆＃39;）

添加了序列号

最后按此等级进行分组，获取最大和最小日期之间的约会。

with tm as
(
    select JobName, CurrentJobStatus, TimeOfChange,
    rank() over (partition by CurrentJobStatus order by TimeOfChange desc) rn
    from jobs
    where CurrentJobStatus in ('Completed', 'Running')
)
select
    rn, DateDiff(minute, min(TimeOfChange), max(TimeOfChange)) as min
from tm
group by rn;

在此处查看：http://rextester.com/NZFWXO47289

Answer 3

这可能很棘手。如果我们假设“Running”和“Completed”完全交织在一起（没有两个“Running”后跟一个“Completed”），那么你可以通过计算每条记录之前或之前的“Running”记录的数量来识别匹配。其余的只是聚合：

select min(case when CurrentJobStatus = 'Running' then TimeOfChange end) as running_time,
       max(case when CurrentJobStatus = 'Completed' then TimeOfChange end) as completed_time,
       datediff(minute, 
                min(case when CurrentJobStatus = 'Running' then TimeOfChange end),
                max(case when CurrentJobStatus = 'Completed' then TimeOfChange end)
                ) as diff_in_minutes
from (select t.*,
             sum(case when CurrentJobStatus = 'Running' then 1 else 0 end) over (order by TimeOfChange) as running_running
      from t
     ) t
group by running_running;

如果假设不成立，并且您可以连续多次“运行”或连续“完成”，那么您应该提出另一个问题，其中包含适当的样本数据和规则说明。

SQL Server根据日期和状态聚合行

3 个答案: