Question

如果“星期日”然后下一个立即日期n + 1一年，如何获得每个月的第n个日期。例如：

 Year=2014
 nth value = 2(2nd of each month)

 Date:
 2014-01-02
 2014-02-03 (since 02 is "Sunday" next Immediate date)
 2014-03-03 (since 02 is "Sunday" next Immediate date)
 2014-04-02
 2014-05-02
 2014-06-02
 2014-07-02 
 and so on...

Answer 1

怎么样，fiddle here

将使用'01'和'28'之间的日期值，我已将参数检查和转换保留为OP。

DECLARE @Year = Char(4);
DECLARE @Day = VarChar(2);

SET @Year = '2014';
SET @Day = '2';

SELECT
    CASE DATENAME(dw, [Date])
        WHEN 'Sunday' THEN DATEADD(d, 1, [Date])
        ELSE [Date]
    END [Date]
  FROM (
    SELECT
        CAST(@Year + '-' + [M] + '-' + @Day AS DATETIME) [Date]
      FROM (
          SELECT '1' [M] UNION ALL
          SELECT '2' UNION ALL
          SELECT '3' UNION ALL
          SELECT '4' UNION ALL
          SELECT '5' UNION ALL
          SELECT '6' UNION ALL
          SELECT '7' UNION ALL
          SELECT '8' UNION ALL
          SELECT '9' UNION ALL
          SELECT '10' UNION ALL
          SELECT '11' UNION ALL
          SELECT '12') [Months]) [RawDays];

Answer 2

这对我有用......！

    Declare @Tracker int
    set @Tracker = 2

    Declare @FromDate datetime
    Declare @ToDate datetime
    set @FromDate = '2014-01-01'
    set @ToDate = '2014-12-31'

    Declare @TrackerTable Table ( Date Datetime)

    ;with DateDifference As
    (
         SELECT @FromDate DateValue
         UNION ALL
         SELECT  DateValue + 1
         FROM    DateDifference   
         WHERE   DateValue + 1 < @ToDate
    )

    Insert into @TrackerTable
        Select 
            Case 
                when datename(dw,DateValue) = 'Sunday' Then DateValue + 1 
                Else DateValue 
                End as Date
            from DateDifference where Day(DateValue) = @Tracker
            OPTION (MAXRECURSION 0) 

    Select * from @TrackerTable

如何获得一年中每个月的第n个日期

2 个答案: