Question

我有一个查询，我真的不知道如何开始。我希望有人可以帮我解决这个问题。我将首先解释表格

我有一个包含四列的设备表：Device_Id，Device_Status，Begin_dt，End_dt 有6种不同的状态，其中3（为简单起见，即状态1,4和5）表示设备是“在线”

此表的一个示例可能是

Id | Status| Begin                   |  End
001|     1 | 2012-09-01 00:00:00.000 | 2012-09-01 01:00:00.000
001|     2 | 2012-09-01 01:00:00.000 | 2012-09-01 01:35:00.000
001|     1 | 2012-09-01 01:35:00.000 | 2012-09-01 02:05:00.000
003|     1 | 2012-09-01 05:00:00.000 | 2012-09-01 07:02:00.000
004|     1 | 2012-09-01 01:00:00.000 | 2012-09-01 01:35:00.000
003|     2 | 2012-09-01 07:02:00.000 | NULL

我的查询需要返回时间总和，所有设备的状态均指示24小时内每小时的“在线”。

所以我的回报应该是

Hour| Online_Time
0   | 5:30:12.11
1   | 3:30:12.11
2   | 4:30:12.11
3   | 5:30:12.11
4   | 6:30:12.11
5   | 4:00:00.00
6   | 1:30:12.11
7   | 3:30:12.11
8   | 4:30:12.11
etc |

因此，对于一天中的每个小时，我可以有超过1小时的在线时间（显然），因为例如，如果我整个小时都有5台设备在线，那么我将有5小时的在线时间。< / p>

这有点复杂，我希望我能很好地解释它，任何有帮助或提出建议的人都非常感激。

-J

with const as (
    select dateadd(hour, 1, cast(cast(getdate() -1 as date) as datetime)) as midnnight
    ),
allhours as (
    select 0 as hour, midnight as timestart, dateadd(hour, 1, timestart) as timeend from const union all
    select 1 as hour, dateadd(hour, 1, midnight), dateadd(hour, 2, midnight) from const union all
    select 2 as hour, dateadd(hour, 2, midnight), dateadd(hour, 3, midnight) from const union all
    select 3 as hour, dateadd(hour, 3, midnight), dateadd(hour, 4, midnight) from const union all
    select 4 as hour, dateadd(hour, 4, midnight), dateadd(hour, 5, midnight) from const union all
    select 5 as hour, dateadd(hour, 5, midnight), dateadd(hour, 6, midnight) from const union all
    select 6 as hour, dateadd(hour, 6, midnight), dateadd(hour, 7, midnight) from const union all
    select 7 as hour, dateadd(hour, 7, midnight), dateadd(hour, 8, midnight) from const union all
    select 8 as hour, dateadd(hour, 8, midnight), dateadd(hour, 9, midnight) from const union all
    select 9 as hour, dateadd(hour, 9, midnight), dateadd(hour, 10, midnight) from const union all
    select 10 as hour, dateadd(hour, 10, midnight), dateadd(hour, 11, midnight) from const union all
    select 11 as hour, dateadd(hour, 11, midnight), dateadd(hour, 12, midnight) from const union all
    select 12 as hour, dateadd(hour, 12, midnight), dateadd(hour, 13, midnight) from const union all
    select 13 as hour, dateadd(hour, 13, midnight), dateadd(hour, 14, midnight) from const union all
    select 14 as hour, dateadd(hour, 14, midnight), dateadd(hour, 15, midnight) from const union all
    select 15 as hour, dateadd(hour, 15, midnight), dateadd(hour, 16, midnight) from const union all
    select 16 as hour, dateadd(hour, 16, midnight), dateadd(hour, 17, midnight) from const union all
    select 17 as hour, dateadd(hour, 17, midnight), dateadd(hour, 18, midnight) from const union all
    select 18 as hour, dateadd(hour, 18, midnight), dateadd(hour, 19, midnight) from const union all
    select 19 as hour, dateadd(hour, 19, midnight), dateadd(hour, 20, midnight) from const union all
    select 20 as hour, dateadd(hour, 20, midnight), dateadd(hour, 21, midnight) from const union all
    select 21 as hour, dateadd(hour, 21, midnight), dateadd(hour, 22, midnight) from const union all
    select 22 as hour, dateadd(hour, 22, midnight), dateadd(hour, 23, midnight) from const union all
    select 23 as hour, dateadd(hour, 23, midnight), dateadd(hour, 24, midnight) from const union all
   ) 
select ah.hour,
   sum(datediff(ms, (case when ah.timestart >= dt.Begin_Dt then timestart else dt.Begin_Dt end),
                    (case when ah.timeend <= dt.End_Dt then ah.timeend else dt.End_Dt end))) as totalms,
   cast(dateadd(ms, sum(datediff(ms, (case when ah.timestart >= dt.Begin_Dt then timestart else dt.Begin_Dt end),
                                 (case when ah.timeend <= dt.End_Dt then ah.timeend else dt.End_Dt end))),0) as time
                                 ) as totalTime
from allhours as ah left outer join
     dataTable as dt
     on ah.timestart< coalesce(dt.End_dt, getdate()) and
        ah.timeend >= dt.Begin_Dt
group by ah.hour
order by ah.hour

这就是我现在所拥有的，我在'）'

上收到错误

select 23 as hour, dateadd(hour, 23, midnight), dateadd(hour, 24, midnight) from const union all
   )  <----- Incorrect syntax near ')'. Expecting SELECT, or '('.
select ah.hour,

Answer 1

这样的事情可能会让你开始。

select  SUM(datediff(second, Begin_dt, End_dt)) as seconds_online
        ,DATEPART(hour, Begin_dt) as [hour]
from table
where device_status in (1,4,5)
group by DATEPART(hour, Begin_dt)

要格式化结果，您可能需要关注此问题：

SQL - Seconds to Day, Hour, Minute, Second

Answer 2

您的问题似乎是时间跨度需要在几小时内分解。所以，你需要从一天中的所有时间开始。然后计算重叠，总结差异（以毫秒为单位）并将所有内容转换回输出的时间。

with const as (
        select dateadd(hour, 1, cast(cast(getdate() -1 as date) as datetime)) as midnight            
    ),
    allhours as (
        select 0 as hour, midnight as timestart, dateadd(hour, 1, midnight) as timeend from const union all
        select 1 as hour, dateadd(hour, 1, midnight), dateadd(hour, 2, midnight) from const union all
        select 2 as hour, dateadd(hour, 2, midnight), dateadd(hour, 3, midnight)  from const union all
        . . .
        select 23 as hour, dateadd(hour, 23, midnight), dateadd(hour, 24, midnight) from const
    )
select ah.hour,
       sum(datediff(ms, (case when ah.timestart >= dt.begin then timestart else dt.begin end),
                        (case when ah.timeend <= dt.end then ah.timeend else dt.end end)
                   ) 
           ) as totalms,
       cast(dateadd(ms, sum(datediff(ms, (case when ah.timestart >= dt.begin then timestart else dt.begin end),
                                     (case when ah.timeend <= dt.end then ah.timeend else dt.end end)
                                    )
                           ),
                     0) as time
           ) as totalTime
from allhours ah left outer join
     DeviceTable dt
     on ah.timestart< coalesce(dt.end, getdate()) and
        ah.timeend >= dt.begin
group by ah.hour
order by ah.hour

此外，为了使其工作，您需要将“begin”和“end”包装在双引号或方括号中。这些是T-SQL中的保留字。你需要更换“......”。额外的线路从3到22小时。

SQL Query返回过去24小时内每小时的值

2 个答案: