Question

我有一张包含id，play，starttime和endtime的表。我想找到每天的总播放时间。我认为查询将类似于以下，但我相信它是不对的。如果我在没有比赛的情况下获得0，但是如果我很难，我也不会介意，这也会非常方便。

Select 
id, 
play,
date,
CASE
    WHEN datediff(day, starttime, endtime) = 0 then sum(totaltime)
END as TimePerDay
from cte where starttime >= '2015-05-30 17:11:34.000'
group by id, playtime, starttime, endtime

我正在寻找

id | play    | Date        | totaltime
1  | hockey  | 05/06/2015  | 0
2  | hockey  | 04/06/2015  | 0
3  | hockey  | 03/06/2015  | 230
4  | hockey  | 02/06/2015  | 10
5  | hockey  | 01/06/2015  | 120

Answer 1

你能试试吗

Select play, cast(starttime as date) as date,
       SUM(datediff(MINUTE, endtime, starttime)) as TimePerDay
from cte
where starttime >= '2015-05-30 17:11:34.000'
group by play, cast(starttime as date)

union


SELECT 'hockey', DATEADD(DAY,number+1,(select min(starttime) from cte)) as date, 0 as TimePerDay
                FROM master..spt_values
                WHERE type = 'P'
                AND DATEADD(DAY,number+1,(select min(starttime) from cte)) < (select max(starttime) from cte)
                and CAST(DATEADD(DAY,number+1,(select min(starttime) from cte)) as date) not in (select cast(starttime as date) from cte)

快递版：

DECLARE @startDate date = (select min(starttime) from cte)
DECLARE @endDate date = (select max(starttime) from cte)


;WITH dates(Date) AS 
(
    SELECT @startdate as Date
    UNION ALL
    SELECT DATEADD(d,1,[Date])
    FROM dates 
    WHERE DATE < @enddate
)
Select play, cast(starttime as date) as date,
       SUM(datediff(MINUTE, endtime, starttime)) as TimePerDay
from cte
where starttime >= '2015-05-30 17:11:34.000'
group by play, cast(starttime as date)

union 

SELECT 'hockey' as play, Date, 0 as TimePerDay
FROM dates
where Date not in (select cast(starttime as date) from cte)

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1 个答案: