Question

我在SQL数据库中有一个表，其中包含有关员工多年工作时间的信息。每个雇员在特定日期可以有多个记录，并且每个雇员的开始日期可以不同。

我试图根据每个员工的第一周来总结他们的每周工作时间。因此，如果该员工于2018年4月17日开始工作，则本周登录的任何时间都将被视为该员工的第1周，而下一周则是第二周，依此类推。对于另一位员工，一周可以在不同的日期/月份/年份等开始。

我的数据包括以下字段：

Sequence_ID：与单个员工有关

Date_European：与员工记录的每个工作日有关，最少为员工首次在公司开始工作的日期

小时数：记录的小时数

我在数据中还有一个Year字段，它是Date_European列的年份。

以下是我尝试过的内容，但我知道它甚至与我需要的格式不符。

select 
    Sequence_ID
    ,DATEPART(week,Date_European) AS Week
    ,DATEPART(year,Date_European) AS Year
    ,SUM([Hours]) AS Weekly_Hours
from [AB_DCU_IP_2018].[dbo].[mytable]
group by 
    Sequence_ID
    ,DATEPART(week,Date_European)
    ,DATEPART(year,Date_European)
order by 
    Sequence_ID
    ,DATEPART(week,Date_European)
    ,DATEPART(year,Date_European)

我尝试创建“周”字段。从上面的代码中，它仅能告诉我日期与特定年份的哪个星期相关。然后，我添加了“年份”列以区分不同的年份，但这再次仅给出了特定的年份。

有什么方法可以按照我想要的格式创建“周”字段？（最早的日期和周围的日期是第1周）。

我试图按功能使用等级和分区，无法使其正常工作。

我一直在寻找解决方案数小时，对您的任何帮助将不胜感激。

谢谢。

编辑：如何创建初始表

CREATE TABLE mytable(Sequence_ID   VARCHAR(6) NOT NULL ,Date_European DATE NOT NULL ,Hours NUMERIC(5,1) NOT NULL);

INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/05/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/06/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/07/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/08/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/09/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/12/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/13/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/14/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/15/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/16/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/19/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/20/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/21/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/22/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/23/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/26/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/27/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/28/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/29/2016',7.3);
INSERT INTO mytable(Sequence_ID,Date_European,Hours) VALUES ('da6Wrw','09/30/2016',7.3);

我想要的结果：

| Sequence_ID | Date_European | DATEPART(week,Date_European) | Hours | Desired_OutCome_Week |
| da6Wrw      | 05/09/2016    |                           37 |   7.3 |                    1 |
| da6Wrw      | 06/09/2016    |                           37 |   7.3 |                    1 |
| da6Wrw      | 07/09/2016    |                           37 |   7.3 |                    1 |
| da6Wrw      | 08/09/2016    |                           37 |   7.3 |                    1 |
| da6Wrw      | 09/09/2016    |                           37 |   7.3 |                    1 |
| da6Wrw      | 12/09/2016    |                           38 |   7.3 |                    2 |
| da6Wrw      | 13/09/2016    |                           38 |   7.3 |                    2 |
| da6Wrw      | 14/09/2016    |                           38 |   7.3 |                    2 |
| da6Wrw      | 15/09/2016    |                           38 |   7.3 |                    2 |
| da6Wrw      | 16/09/2016    |                           38 |   7.3 |                    2 |
| da6Wrw      | 19/09/2016    |                           39 |   7.3 |                    3 |
| da6Wrw      | 20/09/2016    |                           39 |   7.3 |                    3 |
| da6Wrw      | 21/09/2016    |                           39 |   7.3 |                    3 |
| da6Wrw      | 22/09/2016    |                           39 |   7.3 |                    3 |
| da6Wrw      | 23/09/2016    |                           39 |   7.3 |                    3 |
| da6Wrw      | 26/09/2016    |                           40 |   7.3 |                    4 |
| da6Wrw      | 27/09/2016    |                           40 |   7.3 |                    4 |
| da6Wrw      | 28/09/2016    |                           40 |   7.3 |                    4 |
| da6Wrw      | 29/09/2016    |                           40 |   7.3 |                    4 |
| da6Wrw      | 30/09/2016    |                           40 |   7.3 |                    4 |

Answer 1

Set DateFirst 1

select 
    Sequence_ID,
    (datediff(day , DQ.WeekStarted, Date_European) / 7 + 1) EmployeeWeekNumber
    ,SUM([Hours]) AS Weekly_Hours
--into [AB_DCU_IP_2018].[dbo].[Weekly_Work_Hours_Employee]
from [AB_DCU_IP_2018].[dbo].[All_IPower_HR_Assurance_4]
CROSS APPLY (SELECT DATEADD(day, -1 * (datepart(weekday,start_date) % 7), start_date)  AS WeekStarted   
                                    FROM YourTable 
                                    WHERE <condition  to get the start_date you need>
            ) DQ
group by 
    Sequence_ID,
   (datediff(day , DQ.WeekStarted, Date_European) / 7 + 1)
order by 
    Sequence_ID
    ,DATEPART(week,Date_European)
    ,DATEPART(year,Date_European)

Answer 2

这是使用您发布的示例数据的另一种方法。

select mt.Sequence_ID
    , mt.Date_European 
    , DATEPART(week, mt.Date_European) 
    , mt.Hours
    , MyRow.GroupNum
from  mytable mt
join
(
    select WeekNum = DATEPART(week,Date_European) 
        , GroupNum = ROW_NUMBER() over(order by DATEPART(week,Date_European))
    from  mytable
    group by DATEPART(week,Date_European) 

) MyRow on MyRow.WeekNum = DATEPART(week, mt.Date_European)

Answer 3

尝试

select *,rn-1 [Employee_week] from (
select *,dense_RANK() over(Partition by Sequence_ID order by iif(weekly_hours=0,0,week) ) [rn] from (
select 
    Sequence_ID
    ,DATEPART(week,Date_European) AS Week
    ,DATEPART(year,Date_European) AS Year
    ,SUM([Hours]) AS Weekly_Hours
--into [AB_DCU_IP_2018].[dbo].[Weekly_Work_Hours_Employee]
from [AB_DCU_IP_2018].[dbo].[All_IPower_HR_Assurance_4]
group by 
    Sequence_ID
    ,DATEPART(week,Date_European)
    ,DATEPART(year,Date_European)
order by 
    Sequence_ID
    ,DATEPART(week,Date_European)
    ,DATEPART(year,Date_European))a)a
    where rn = 2

这将为您提供每个员工在第一周的工作时间，使用rn> 2来获取剩余的几周

Answer 4

我实际上找到了一种更简单的方法来计算使用DENSE_Rank函数的员工的工作周数。

如果有人出现类似问题，我将在下面列出。我已经注释掉DATEPART部分，因为我只是使用这些列作为检查以确保其正常工作：

select 
     Sequence_ID
    ,Date_European
    --,DATEPART(week,Date_European) AS Week
    --,DATEPART(year,Date_European) AS Year
    ,DENSE_RANK() OVER (PARTITION BY Sequence_ID ORDER BY DATEPART(year,Date_European), DATEPART(week,Date_European) asc) AS EmployeeWeekNumber
    ,Hours
from [AB_DCU_IP_2018].[dbo].[All_IPower_HR_Assurance_4]
order by 
     Sequence_ID 
    ,Date_European
    --,DATEPART(week,Date_European)
    --,DATEPART(year,Date_European)

几年中按周计算的SQL SUM小时

4 个答案: