[英]SQL SUM hours by week across number of years
我在SQL數據庫中有一個表,其中包含有關員工多年工作時間的信息。 每個雇員在特定日期可以有多個記錄,並且每個雇員的開始日期可以不同。
我試圖根據每個員工的第一周來總結他們的每周工作時間。 因此,如果該員工於2018年4月17日開始工作,則本周登錄的任何時間都將被視為該員工的第1周,而下一周則是第二周,以此類推。對於另一位員工,一周可以在不同的日期/月份/年等
我的數據包括以下字段:
Sequence_ID:與單個員工有關
Date_European:與員工記錄的每個小時相關的日期,其中最少的時間是該員工在公司開始的首次日期
小時:記錄的小時數
我在數據中還有一個Year字段,它是Date_European列的年份。
以下是我嘗試過的內容,但我知道它甚至與我需要的格式不符。
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)
我嘗試創建“周”字段。 從上面的代碼中,它僅能告訴我日期與特定年份的哪個星期相關。 然后,我添加了“年份”列以區分不同的年份,但這再次僅給出了特定的年份。
有什么方法可以按照我想要的格式創建“周”字段? (最早的日期和周圍的日期是第1周)。
我試圖按功能使用等級和分區,無法使其正常工作。
任何幫助將不勝感激,因為我一直在尋找解決方案數小時。
提前致謝。
編輯:如何創建初始表
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);
我想要的結果是什么:
| 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 |
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)
這是使用您發布的樣本數據的另一種方法。
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)
嘗試這個
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
這將為您提供每個員工在第一周工作的時間,使用rn> 2來獲取剩余的幾周
實際上,我發現了一種更簡單的方法來計算使用DENSE_Rank函數的員工的工作周數。
如果有人出現類似問題,我將在下面列出。 我已將DATEPART部分注釋掉,因為我只是使用這些列作為檢查以確保其正常工作:
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)
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