Question

我有一个如下表，我需要按周（系统日期）分组

 tran_date      Hours   Tech_ID Tech_Name
02/02/2019 16:30    1   310091  Nasser Ahmed Muhammad
02/02/2019 17:30    8   310094  Ajish Babu Parambath
03/02/2019 08:30    8   310380  Nishad Mulayath Mohanan
04/02/2019 08:30    8   310380  Nishad Mulayath Mohanan
04/02/2019 08:30    8   310094  Ajish Babu Parambath
05/02/2019 08:30    1   310091  Nasser Ahmed Muhammad
06/02/2019 08:30    1   310380  Nishad Mulayath Mohanan
06/02/2019 08:30    5   310094  Ajish Babu Parambath
09/02/2019 09:30    3   310091  Nasser Ahmed Muhammad
09/02/2019 09:30    3   310094  Ajish Babu Parambath
09/02/2019 11:00    1   310380  Nishad Mulayath Mohanan
09/02/2019 13:30    1   310380  Nishad Mulayath Mohanan
09/02/2019 13:30    1   310091  Nasser Ahmed Muhammad
09/02/2019 14:30    1   310094  Ajish Babu Parambath
09/02/2019 16:30    2   310094  Ajish Babu Parambath
10/02/2019 08:30    1   310094  Ajish Babu Parambath
10/02/2019 08:30    2   310380  Nishad Mulayath Mohanan
10/02/2019 09:30    1   310094  Ajish Babu Parambath
10/02/2019 10:30    2   310091  Nasser Ahmed Muhammad
10/02/2019 13:30    1   310380  Nishad Mulayath Mohanan
10/02/2019 13:30    2   310094  Ajish Babu Parambath
10/02/2019 14:30    1   310380  Nishad Mulayath Mohanan
10/02/2019 15:30    1   310094  Ajish Babu Parambath
10/02/2019 16:30    1   310094  Ajish Babu Parambath
13/02/2019 08:30    4   310380  Nishad Mulayath Mohanan
14/02/2019 13:30    1   310091  Nasser Ahmed Muhammad
14/02/2019 14:30    0.5 310091  Nasser Ahmed Muhammad

我期望得到如下所示的结果。

tran_week              Total
2/1/2019 - 2/7/2019    40
2/8/2019 - 2/14/2019   29.5

Answer 1

只需尝试此代码

=>  /aaa
=>  /bbb
=>  /ccc

Answer 2

您可以尝试以下操作。

;with cte as 
(
 select cast(tran_date as date) tran_date, Hours
 from @mytable
)

SELECT Convert(varchar, WS, 101) + '-' + Convert(varchar, WE, 101) AS trans_week ,SUM(HOURS) Total FROM
(
    SELECT      DATEADD(dd, -(DATEPART(dw, tran_date)-1), tran_date) WS, DATEADD(dd, 7-(DATEPART(dw, tran_date)), tran_date) WE
            ,   Hours           
    FROM        cte
)
T
GROUP BY WS,WE
order by ws

Online Demo

注意：如果您想更改一周的第一天，则需要像下面这样先进行设置。

SET DATEFIRST 6

SET DATEFIRST

Answer 3

如果您通过选择的编程语言生成查询的select部分的字符串，则以下代码将起作用，因此将能够计算小时数：

    select 
      sum (case when t.tran_date between  range1_1 and  range1_2 then t.hours else 0 end) as range1,
      sum (case when t.tran_date between  range2_1 and  range2_2 then t.hours else 0 end) as range2,
      sum (case when t.tran_date between  range3_1 and  range3_2 then t.hours else 0 end) as range3,
      sum (case when t.tran_date between  range3_1 and  range4_2 then t.hours else 0 end) as range4,
  ...
  from your_table as t

在SQL中按1-7系统日期分组

3 个答案: