Question

SELECT DISTINCT (DATEDIFF(DAY, '2014-06-01', '2014-06-30') + 1) - DATEDIFF(WEEK, '2014-06-01', '2014-06-30') * 2 - (CASE WHEN DATEPART(WEEKDAY, '2014-06-01') = 5 THEN 1 ELSE 0 END) - (CASE WHEN DATEPART(WEEKDAY, '2014-06-30') = 6 THEN 1 ELSE 0 END) AS TotalWorkingDays, COUNT(DISTINCT DATEADD(d, 0,DATEDIFF(d, 0, CHECKINOUT.CHECKTIME))) AS workingdays,
 USERINFO.NAME,
case when  '08:00:00.000' > cast(convert(char(8),min(checktime),8) as datetime) THEN NULL
ELSE 
cast(dateadd(minute,datediff(minute,'08:00:00.000',min(checktime)),'00:00')AS dateTIME) END as LATEIN ,
case when '17:00:00.000' < cast(convert(char(8),max(checktime),8) as datetime) THEN NULL 
ELSE
cast(dateadd(minute,datediff(minute,max(checktime),'17:00:00.000'),'00:00')AS dateTIME)END as LATEOUT     
FROM   USERINFO left JOIN CHECKINOUT ON CHECKINOUT.USERID = USERINFO.USERID 
left join departments on deptid = userinfo.DEFAULTDEPTID
WHERE(DEPARTMENTS.DEPTNAME = 'GEN/SUP-TBL') AND (CHECKINOUT.CHECKTIME >='2014-06-01') AND (CHECKINOUT.CHECKTIME <='2014-06-30') and name='Gokul Gopalakrishnan'
GROUP BY USERINFO.NAME,checktime

这是我的查询，它给出了以下输出：

TotalWorkingDays    workingdays NAMER   LATEIN                          LATEOUT
22                  1           abc      NULL                   1785-07-17 09:24:00.000
22                  1           abc      NULL                   1785-07-18 00:15:00.000
22                  1           abc      NULL                   1785-07-31 00:30:00.000
22                  1           abc    2014-06-01 00:39:00.000  1785-08-02 01:00:00.000
22                  1           abc    2014-06-01 09:14:00.000     NULL

但现在我想要输出如下：

TotalWorkingDays    workingdays NAME    TOTAL LATEIN        TOTAL LATEOUT
22                    5         abc     09:53:00.000       10:69:00.000

那我该如何在sql server中执行此操作？

Answer 1

您需要做的是将第一个选择的结果插入到声明的表中，然后对声明的表的列求和。在我的代码中存在1785年的问题（这就是我使用2000代替的原因），但我正在努力：）

DECLARE @unSummedResults TABLE (totalWorkingDays int,  workingdays int, name    nvarchar(10), latein smalldatetime, lateout smalldatetime)

insert into @unSummedResults (totalWorkingDays,  workingdays, name, latein, lateout) values (22, 1, 'abc', NULL, '2000-07-17 09:24:00.000')
insert into @unSummedResults (totalWorkingDays,  workingdays, name, latein, lateout) values (22, 1, 'abc', NULL, '2000-07-18 00:15:00.000')
insert into @unSummedResults (totalWorkingDays,  workingdays, name, latein, lateout) values (22, 1, 'abc', NULL, ' 2000-07-31 00:30:00.000')
insert into @unSummedResults (totalWorkingDays,  workingdays, name, latein, lateout) values (22, 1, 'abc', '2014-06-01 00:39:00.000', '2000-08-02 01:00:00.000')
insert into @unSummedResults (totalWorkingDays,  workingdays, name, latein, lateout) values (22, 1, 'abc', '2014-06-01 09:14:00.000', NULL)

select distinct unSummedResults.totalWorkingDays, sum(workingdays), name, 
    convert(varchar(5), sum(datepart(hh, latein))) + ':' + 
    convert(varchar(5), sum(datepart(n, latein))) + ':' + 
    convert(varchar(5), sum(datepart(s, latein))), 
    convert(varchar(5), sum(datepart(hh, lateout))) + ':' + 
    convert(varchar(5), sum(datepart(n, lateout))) + ':' + 
    convert(varchar(5), sum(datepart(s, lateout)))
from @unSummedResults as  unSummedResults
group by totalWorkingDays, name

好运！

sql server中以小时/分钟为单位的时间总和

1 个答案: