如何获得一年中每个月的第n个日期

Question

How to get the nth date of each month if "sunday" then that next Immediate date n+1 for a year. for example :

 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

how about this, fiddle here

Will work with day values between '01' and '28' , I've left parameter checking and conversion to the 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

This worked for me...!

    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