如何在SQL Server中獲取第2個和第4個星期六日期列表？

Question

我幾乎是編寫SQL查詢的新手。 在SQL Server的上下文中，如何獲取2016年第2個和第4個星期六日期列表 ？

Answer 1

完成作爲派生表只是爲了顯示邏輯，但如果你喜歡，你可以減少：

select * 
from (
    select d2016, 
      datename(weekday, d2016) as wkdy, 
      row_number() over (partition by datepart(month, d2016), datename(weekday, d2016) order by d2016) as rn_dy_mth 
    from (
     select dateadd(day, rn, cast('2016-01-01' as date)) as d2016 
     from (
      select row_number() over(order by object_id) - 1 as rn 
      from sys.columns 
      ) as rn 
     ) as dy 
    ) as dy_mth 
where rn_dy_mth in (2, 4) 
    and wkdy = 'Saturday' 
order by d2016 

Answer 2

--DEFINE LIMITS FOR DAY 
DECLARE @TODATE DATETIME, @FROMDATE DATETIME 
SET @FROMDATE ='2010-01-01' 
SET @TODATE = '2017-12-31' 

;WITH DATESEQUENCE([DATE]) AS 
(
    SELECT @FROMDATE AS [DATE] 
     UNION ALL 
    SELECT DATEADD(DAY, 1, [DATE]) 
     FROM DATESEQUENCE 
     WHERE DATE < @TODATE 
    ) 
, DATESATURDAY AS 
(SELECT CAST(CAST(YEAR([DATE]) AS VARCHAR)+ 
     (CASE WHEN DATEPART(M,[DATE])<=9 THEN '0'+CAST(DATEPART(M,[DATE]) AS VARCHAR) 
     ELSE CAST(DATEPART(M,[DATE]) AS VARCHAR) END) AS NUMERIC) AS MONTH_ID 

     ,CONVERT(VARCHAR,[DATE],106) AS DAY_DESC 

     ,UPPER(DATENAME(DW,[DATE]))AS DAY_NAME 
FROM DATESEQUENCE) 
,SECOND_FOURTH_SATURDAY AS 
(SELECT * 
,ROW_NUMBER() OVER (PARTITION BY MONTH_ID ORDER BY DAY_NAME) FALL_IN 
FROM DATESATURDAY 
WHERE DAY_NAME='SATURDAY') 
SELECT * FROM SECOND_FOURTH_SATURDAY 
WHERE FALL_IN IN(2,4) 
OPTION (MAXRECURSION 10000)