查找最後一個星期日

Question

如何在sql 2000中查找一個月的最後一個星期日？

Answer 1

SELECT 
DATEADD(day,DATEDIFF(day,'19000107',DATEADD(month,DATEDIFF(MONTH,0,GETDATE() /*YourValuehere*/),30))/7*7,'19000107') 

編輯：正確的，最後，工作從我的同事回答。

Answer 2

天哪，這是醜陋的，但這裏有雲：

DECLARE @dtDate DATETIME 
SET @dtDate = '2009-11-05' 

SELECT DATEADD(dd, -1*(DATEPART(dw, DateAdd(day, -1, DateAdd(month, DateDiff(month, 0, @dtDate)+1, 0)))-1), 
      DateAdd(day, -1, DateAdd(month, DateDiff(month, 0, @dtDate)+1, 0))) 

Answer 3

首先建立一個理貨表。 http://www.sqlservercentral.com/articles/T-SQL/62867/ 然後得到你想要的東西..

http://www.sqlservercentral.com/Forums/Topic515226-1291-1.aspx

DECLARE @DateStart DATETIME, 
     @DateEnd DATETIME 

SELECT @DateStart = '20080131', 
     @DateEnd = '20101201' 

SELECT DATEADD(wk,DATEDIFF(wk,6,DATEADD(mm,DATEDIFF(mm,-1,DATEADD(mm,t.N-1,@DateStart)),-1)),6) 
    FROM dbo.Tally t 
    WHERE t.N <= DATEDIFF(mm,@DateStart,@DateEnd) 

Answer 4

DECLARE @LastDateOfMonth smalldatetime 
SELECT @LastDateOfMonth = DATEADD(month, DATEDIFF(month, -1, GETDATE()), 0) -1 
Select DATEADD(dd,-(CASE WHEN DATEPART(weekday,@LastDateOfMonth) = 1 THEN 0 ELSE DATEPART(weekday,@LastDateOfMonth) - 1 END),@LastDateOfMonth) 

Answer 5

的另一種方法，從數據倉庫實踐借來的。創建一個日期維度表並預加載10年左右。

TABLE dimDate (DateKey, FullDate, Day, Month, Year, DayOfWeek, 
       DayInEpoch, MonthName, LastDayInMonthIndicator, many more..) 

以填補在dimDate最簡單的方法是一個下午花與Excel，然後從那裏導入到數據庫中。一半體面dimDate表有50多列 - 任何你想知道約會的東西。

有了這個地方，問題就變成這樣的：

SELECT max(FullDate) 
FROM dimDate 
WHERE DayOfWeek = 'Sunday' 
     AND Month = 11 
     AND Year = 2009; 

從本質上講，所有日期相關的查詢變得更簡單。

Answer 6

select next_day(last_day(sysdate)-7, 'Sunday') from dual 

Answer 7

select dateadd(day,1-datepart(dw, getdate()), getdate()) 

Answer 8

下週日在SQL，無論哪一天是一週的第一天：返回2011-01-02 23：59：59.000在22日 - 12月2010：

select DateADD(ss, -1, DATEADD(week, DATEDIFF(week, 0, getdate()), 14)) 

Answer 9

我找到一些難以理解的解決方案，所以這裏是我的帶變量的版本來解釋步驟。

ALTER FUNCTION dbo.fn_LastSundayInMonth 
(
    @StartDate DATETIME 
,@RequiredDayOfWeek INT /* 1= Sunday */ 
) 
RETURNS DATETIME 
AS 
/* 
A detailed step by step way to get the answer... 

SELECT dbo.fn_LastSundayInMonth(getdate()-31,1) 
SELECT dbo.fn_LastSundayInMonth(getdate()-31,2) 
SELECT dbo.fn_LastSundayInMonth(getdate()-31,3) 
SELECT dbo.fn_LastSundayInMonth(getdate()-31,4) 
SELECT dbo.fn_LastSundayInMonth(getdate()-31,5) 
SELECT dbo.fn_LastSundayInMonth(getdate()-31,6) 
SELECT dbo.fn_LastSundayInMonth(getdate()-31,7) 
*/ 
BEGIN 
    DECLARE @MonthsSince1900 INTEGER 
    DECLARE @NextMonth INTEGER 
    DECLARE @DaysToSubtract INTEGER 
    DECLARE @FirstDayOfNextMonth DATETIME 
    DECLARE @LastDayOfMonthDayOfWeek INTEGER 
    DECLARE @LastDayOfMonth DATETIME 
    DECLARE @ReturnValue DATETIME 

    SET @MonthsSince1900=DateDiff(month, 0, @StartDate) 
    SET @NextMonth=@MonthsSince1900+1 
    SET @FirstDayOfNextMonth = DateAdd(month,@NextMonth, 0) 
    SET @LastDayOfMonth = DateAdd(day, -1, @FirstDayOfNextMonth) 

    SET @ReturnValue = @LastDayOfMonth 

    WHILE DATEPART(dw, @ReturnValue) <> @RequiredDayOfWeek 
     BEGIN 
      SET @ReturnValue = DATEADD(DAY,-1, @ReturnValue) 
     END 

    RETURN @ReturnValue 
END