SQL Server - 給定週數，在一週內獲取第一個日期？

Question

我有一個查詢（用於bug tracker.net），它按周狀態計算每週的bug數量。但是，該查詢返回的週數，我真正想要的是一週的第一天

select datepart(wk, DateAdd(day, 0, DateDiff(day, 0, bg_reported_date))) 
     as [week], bg_status , st_name as [status], count(*) as [count] 
    from bugs inner join statuses on bg_status = st_id 
group by datepart(wk, DateAdd(day, 0, DateDiff(day, 0, bg_reported_date))), 
      bg_status, st_name 
order by [week], bg_status 

是得到週數的部分是

datepart(wk, DateAdd(day, 0, DateDiff(day, 0, bg_reported_date))) as [week] 

它返回的輸出：

week  bg_status status          count 
----------- ----------- --------------------------------------------- ------ 
22   1   new           1 
22   5   closed          32 

但倒不如說每個星期的第一天，如01-01-2010，那麼2010年8月1日，等

問題不是How do you get the "week start date" and "week end date" from week number in SQL Server?重複（回答說如何從一個日期從一週數量得到周開始不）

不是Calculate date from week number重複（問題問的C＃）

不是Get first date of week from provided date重複（問題問的JavaScript）

我沒有搜索，但沒有找到這個問題的回答爲SQL Server（2010，如果它事項）

Answer 1

如果你以正確的方式想想，答案SO 1267126可以應用到你的問題。

每個bug報告日期，你在組中映射到同一周。因此，根據定義，這些錯誤日期中的每一個還必須映射到同一周的開始。所以，你在錯誤報告日期以及週數計算上運行「從給定日期算起的一週的開始」計算，並且通過兩個（適度可怕的）表達式來分組，並最終得到你所尋求的答案。

SELECT DATEPART(wk, DATEADD(day, 0, DATEDIFF(d, 0, bg_reported_date))) [week], 
     DATEADD(dd, -(DATEPART(dw, bg_reported_date)-1), bg_reported_date) 
     AS [weekstart], bg_status, st_name AS [status], COUNT(*) AS [count] 
    FROM bugs INNER JOIN statuses ON bg_status = st_id 
GROUP BY DATEPART(wk, DATEADD(day, 0, DATEDIFF(day, 0, bg_reported_date))), 
     DATEADD(dd, -(DATEPART(dw, bg_reported_date)-1), bg_reported_date), 
     bg_status, st_name 
ORDER BY [week], bg_status 

---

由於bg_reported_date是DATETIME（見註釋;它包括時間分量），有必要確定該周開始之前將它轉換爲DATE（但週數expression不需要投，及本週開始表達的一部分「星期幾」不需要投其一）：

SELECT DATEPART(wk, DATEADD(day, 0, DATEDIFF(d, 0, bg_reported_date))) [week], 
     DATEADD(dd, -(DATEPART(dw, bg_reported_date)-1), 
       CAST(bg_reported_date AS DATE)) AS [weekstart], 
     bg_status, st_name AS [status], COUNT(*) AS [count] 
    FROM bugs INNER JOIN statuses ON bg_status = st_id 
GROUP BY DATEPART(wk, DATEADD(day, 0, DATEDIFF(day, 0, bg_reported_date))), 
     DATEADD(dd, -(DATEPART(dw, bg_reported_date)-1), 
       CAST(bg_reported_date AS DATE), 
     bg_status, st_name 
ORDER BY [week], bg_status 

注：未經測試的代碼！

Answer 2

我意識到這是一個非常古老的線程，但「在給定週數的情況下在一週內獲取第一個日期」正是我想要做的，而且我沒有實際的日期來處理，因此接受的答案不會爲我工作。我想我會爲後人發表我的解決方案。請注意，我懷疑不同的文化設置可能會破壞這一點，所以在使用之前進行測試。

我的答案是從this一開始構建的。

讓我們假設你知道一個星期的數字和一年，並且你想得到那一年那一週的開始和結束日期。這是我有什麼：

--These 2 "declared" variables would be passed in somehow 
declare @WeekNumber int = DATEPART(wk, GETDATE()) 
declare @ForYear int = YEAR(GETDATE())-1 

--Since we don't have a raw date to work with, I figured I could just start with 
--Jan 1 of that year. I'll store that date in a cte here, but if you are doing this 
--in a stored proc or function, it would make much more sense to use another @variable 
;with x as 
(
    --this method works in SQL 2008: 
    SELECT CONVERT(DateTime, ('1/1/' + CONVERT(varchar, @ForYear))) as Jan1ForSelectedYear 
    --If you are using 2014 or higher, you can use this instead: 
    --DATETIME2FROMPARTS(@ForYear, 1, 1, 0,0,0,0,0) 
) 
--Now that we have a date to work with, we'll just add the number of weeks to that date 
--That will bring us to the right week number of the given year. 
--Once we have THAT date, we can get the beginning and ending of that week 
--Sorry to make you scroll, but I think this is easier to see what is going on this way 
SELECT CONVERT(varchar(50), DateAdd(wk, (@WeekNumber - 1), (DATEADD(dd, @@DATEFIRST - DATEPART(dw, x.Jan1ForSelectedYear) - 6, x.Jan1ForSelectedYear))), 101) as FirstDayOfWeekXForSelectedYear, 
     CONVERT(varchar(50), DateAdd(wk, (@WeekNumber - 1), (DATEADD(dd, @@DATEFIRST - DATEPART(dw, x.Jan1ForSelectedYear) , x.Jan1ForSelectedYear))), 101) as LastDayOfWeekXForSelectedYear 
FROM x