Question

sql服务器按周总和分组

我有下面的代码，时间表。下面是时间表列

下面是用户表列用户名，名字，姓氏

下面是时间表的记录

timeid,shiftdate,starttime,endtime,userid
1,1st Jun 2019,1:00,10:00,1
2,2st Jun 2019,1:00,10:00,2
3,3rd Jun 2019,4:00,11:00,1
10,11th Jun 2019,4:00,11:00,1
14,11th Jun 2019,4:00,11:00,1
19,11th Jun 2019,4:00,11:00,1

我的问题是我想生成每周总报告像下面这样

Userid | 3rd June to 9th June | 10th to 16 June
1              10                     20
2               5                      5

我尝试了以下代码

 select sum(datediff(hour,startdate,enddate),userid,shiftdate from time 
group by userid,shiftdate

我每天都在给我关于用户的信息。但是我需要我要在每周的第一天到每周的最后一天之间有小时数。

我每天都在给我关于用户的信息。但是我需要我需要在每周的第一天到每周的最后一天之间有小时数。你能帮忙吗

Answer 1

这可能会帮助

DECLARE @TimeTable TABLE(
timeid INT,shiftdate DATETIME ,starttime TIME,endtime TIME ,userid INT
)
INSERT INTO @TimeTable(timeid,shiftdate,starttime,endtime,userid) VALUES(1,'1 Jun 2019','1:00','10:00',1)
INSERT INTO @TimeTable(timeid,shiftdate,starttime,endtime,userid) VALUES(2,'2 Jun 2019','1:00','10:00',2)
INSERT INTO @TimeTable(timeid,shiftdate,starttime,endtime,userid) VALUES(2,'3 Jun 2019','1:00','10:00',2)
INSERT INTO @TimeTable(timeid,shiftdate,starttime,endtime,userid) VALUES(3,'3 Jun 2019','4:00','11:00',1)
INSERT INTO @TimeTable(timeid,shiftdate,starttime,endtime,userid) VALUES(10,'11 Jun 2019','4:00','11:00',1)
INSERT INTO @TimeTable(timeid,shiftdate,starttime,endtime,userid) VALUES(14,'11 Jun 2019','4:00','11:00',3)
INSERT INTO @TimeTable(timeid,shiftdate,starttime,endtime,userid) VALUES(19,'11 Jun 2019','4:00','11:00',3)
INSERT INTO @TimeTable(timeid,shiftdate,starttime,endtime,userid) VALUES(2,'11 Jun 2019','1:00','10:00',2)
INSERT INTO @TimeTable(timeid,shiftdate,starttime,endtime,userid) VALUES(10,'19 Jun 2019','4:00','11:00',1)
INSERT INTO @TimeTable(timeid,shiftdate,starttime,endtime,userid) VALUES(10,'20 Jun 2019','4:00','11:00',1)

;WITH WeeklyTotal
AS(
SELECT * 
      ,DATEDIFF(HOUR,starttime,endtime) totalHours
      ,DATEPART(WEEK, shiftdate) - DATEPART(WEEK, DATEADD(MM, DATEDIFF(MM,0,shiftdate), 0))+ 1  AS weekOfMonth
FROM @TimeTable
)
SELECT * FROM (
SELECT userid,weekOfMonth,totalHours
FROM WeeklyTotal
) aaa
pivot
(
  SUM(totalHours)
  for weekOfMonth in ([1], [2], [3],[4],[5],[6])
) piv;

如何使SQL按周总计

1 个答案: