Question

我试图获得16小时时间间隔的记录数。以下是我现在使用的代码。

;With Cte_hours as ( --hours generation
    Select top(6) hr = (Row_number() over (order by (Select NULL))-1)*4 from master..spt_values 
), cte2 as ( --getting range
    Select DateAdd(HH, c.hr, Convert(datetime,d.dts) ) as Dts_Start, DateAdd(MS, -2, DateAdd(HH, c.hr+ 4, Convert(datetime,d.dts) ) ) Dts_end
    from (select distinct convert(date, dt) as dts from TEST2 ) d
    cross apply Cte_hours c
) --actual query
Select c2.Dts_Start as DT, Sum(case when t.Dt is not null then 1 else 0 end) No_of_records,LD_VOY_N,LD_VSL_M
from cte2 c2
Left Join TEST2 t
on t.Dt between c2.Dts_Start and c2.Dts_end
group by c2.Dts_Start,LD_VOY_N,LD_VSL_M
order by LD_VOY_N, LD_VSL_M, Dts_Start ASC

此代码能够根据4,6和12小时间隔计算我拥有的记录数。但是，如果我尝试根据16小时间隔进行计数，它就会以某种方式起作用。下面是我用于16小时间隔的代码和输出。

;With Cte_hours as ( --hours generation
    Select top(6) hr = (Row_number() over (order by (Select NULL))-1)*16 from master..spt_values 
), cte2 as ( --getting range
    Select DateAdd(HH, c.hr, Convert(datetime,d.dts) ) as Dts_Start, DateAdd(MS, -2, DateAdd(HH, c.hr+ 16, Convert(datetime,d.dts) ) ) Dts_end
    from (select distinct convert(date, dt) as dts from TEST2 ) d
    cross apply Cte_hours c
) --actual query
Select c2.Dts_Start as DT, Sum(case when t.Dt is not null then 1 else 0 end) No_of_records,LD_VOY_N,LD_VSL_M
from cte2 c2
Left Join TEST2 t
on t.Dt between c2.Dts_Start and c2.Dts_end
group by c2.Dts_Start,LD_VOY_N,LD_VSL_M
order by LD_VOY_N, LD_VSL_M, Dts_Start ASC

结果：

DT                          No_of_records  LD_VOY_N      LD_VSL_M
2017-05-05 16:00:00.000     14             0002W         pqo emzmnwp
2017-05-06 00:00:00.000     14             0002W         pqo emzmnwp
2017-05-06 08:00:00.000     12             0002W         pqo emzmnwp
2017-05-06 16:00:00.000     12             0002W         pqo emzmnwp
2017-05-01 16:00:00.000     1              0007E         omq ynzmeoyn
2017-05-02 00:00:00.000     1              0007E         omq ynzmeoyn

这也需要8小时的时间。你们有谁知道为什么？

分组和计数16小时的时间间隔不起作用

0 个答案: