在 SQL Server 中仅获取每月的周六和周日日期

Question

我能否找到特定月份的星期六和星期日所在的日期？例如，考虑 2017 年 1 月的月份。以下日期为周末：

 7/1/2017 - Saturday
14/1/2017 - Saturday
21/1/2017 - Saturday
28/1/2017 - Saturday 
 1/1/2017 - Sunday
 8/1/2017 - Sunday    
15/1/2017 - Sunday
22/1/2017 - Sunday
29/1/2017 - Sunday

我想要一个 SQL Server 查询，这样当我将月份和年份作为输入传递时，我应该返回所有上述日期（仅限周六和周日的日期）作为输出

我不希望使用任何用户定义的函数并希望在单个 SELECT 语句中完成它

Answer 1

注意：正如 cmets 中的另一位用户已经指出的，此查询取决于您的服务器设置，即 DATEFIRST。如果您因为设置不同需要更改查询，请告诉我，我可以为您更改。

使用 CTE 作为虚拟数据...

/* Ignore this part...*/
WITH CTE AS
(
    SELECT CAST('01/01/2017' AS DATE) AS [Date]
        UNION ALL
    SELECT DATEADD(DAY,1,[Date])
    FROM CTE
    WHERE DATE <= '12/31/2017'
)
/*Your actual SELECT statement would look like this, from your own table of course*/
SELECT 
     [Date]
    ,CASE DATEPART(dw,[Date])
        WHEN 1 THEN 'Sunday'
        WHEN 2 THEN 'Monday'
        WHEN 3 THEN 'Tuesday'
        WHEN 4 THEN 'Wednesday'
        WHEN 5 THEN 'Thursday'
        WHEN 6 THEN 'Friday'
        WHEN 7 THEN 'Saturday'
    END
FROM CTE
WHERE DATEPART(dw,[Date]) IN (1,7)
AND MONTH([Date]) = 12--<month>
AND YEAR([Date]) = 2017--<year>
OPTION (MAXRECURSION 0) -- You won't need this line if you're querying a real table
;

如果运行它对你有用，那么你的真实查询可能看起来像这样：

SELECT 
     [Date]
    ,CASE DATEPART(dw,[Date])
        WHEN 1 THEN 'Sunday'
        WHEN 2 THEN 'Monday'
        WHEN 3 THEN 'Tuesday'
        WHEN 4 THEN 'Wednesday'
        WHEN 5 THEN 'Thursday'
        WHEN 6 THEN 'Friday'
        WHEN 7 THEN 'Saturday'
    END
FROM < the table you want >
WHERE DATEPART(dw,[Date]) IN (1,7) -- Only Sundays and Saturdays
AND MONTH([Date]) = < the month you want >
AND YEAR([Date]) = < the year you want >
;

如果您想生成数据，那么 CTE 就是要走的路。如果你传递参数，它看起来像这样：

DECLARE
     @MONTH INT
    ,@YEAR INT
;

SET @MONTH = 1;
SET @YEAR = 2017;

WITH CTE AS
(
    SELECT CAST(CAST(@MONTH AS VARCHAR(2)) + '/01/' + CAST(@YEAR AS VARCHAR(4)) AS [Date]) AS DATE
        UNION ALL
    SELECT DATEADD(DAY,1,[Date])
    FROM CTE
    WHERE DATE <= CAST(@MONTH AS VARCHAR(2)) + 
        CASE 
            WHEN @MONTH IN (9,4,6,11) 
                THEN '/30/'
            WHEN @MONTH IN (1,3,5,7,8,10,12) 
                THEN '/31/'
            WHEN @MONTH = 2 AND @YEAR/4.00 = @YEAR/4 
                THEN '/29/'
            ELSE '/28/'
        END
        + CAST(@YEAR AS VARCHAR(4))
)
SELECT 
     [Date]
    ,CASE DATEPART(dw,[Date])
        WHEN 1 THEN 'Sunday'
        WHEN 2 THEN 'Monday'
        WHEN 3 THEN 'Tuesday'
        WHEN 4 THEN 'Wednesday'
        WHEN 5 THEN 'Thursday'
        WHEN 6 THEN 'Friday'
        WHEN 7 THEN 'Saturday'
    END
FROM CTE
WHERE DATEPART(dw,[Date]) IN (1,7)
OPTION (MAXRECURSION 0)
;

Answer 2

请试试这个。

DECLARE @Year AS INT=2017,
@Month AS INT=3,
@FirstDateOfYear DATETIME,
@LastDateOfYear DATETIME

SELECT @FirstDateOfYear = DATEADD(yyyy, @Year - 1900, 0)
SELECT @LastDateOfYear = DATEADD(yyyy, @Year - 1900 + 1, 0)
-- Creating Query to Prepare Year Data
;WITH cte AS (
SELECT 1 AS DayID,
@FirstDateOfYear AS FromDate,
DATENAME(dw, @FirstDateOfYear) AS Dayname
UNION ALL
SELECT cte.DayID + 1 AS DayID,
DATEADD(d, 1 ,cte.FromDate),
DATENAME(dw, DATEADD(d, 1 ,cte.FromDate)) AS Dayname
FROM cte
WHERE DATEADD(d,1,cte.FromDate) < @LastDateOfYear
)
SELECT FromDate AS Date, Dayname
FROM CTE
WHERE DayName IN ('Saturday','Sunday') and month(FromDate) = @Month 
OPTION (MaxRecursion 370)

Answer 3

这应该可以解决问题：

DECLARE @month date = '2017-01-01'

SET @month = dateadd(month, datediff(month, 0, @month), 0)

;WITH CTE as
(
  SELECT 0 x
  FROM (values(1),(1),(1),(1),(1),(1)) x(n)
),
CTE2 as
(
  SELECT 
    top(day(eomonth(@month))) 
    -- use this syntax for sqlserver 2008
    -- top(datediff(d, @month,dateadd(month,1,@month))) 
    cast(dateadd(d, row_number()over(order by(select 1))-1,@month) as date) cDate
  FROM CTE CROSS JOIN CTE C2
)
SELECT
  cDate, 
  datename(weekday, cDate) Weekday
FROM CTE2
WHERE 
  datediff(d,0,cDate)%7 > 4

Fiddle

Answer 4

来自https://www.sqlservercentral.com/articles/finding-the-correct-weekday-regardless-of-datefirst，您只需：

(DATEPART(dw, @your_date) + @@DATEFIRST) % 7 不在 2 和 6 之间

Answer 5

根据要求，单个select，语言中立，dateFirst 中立，几乎 SQL 版本中立：

declare @OneDate datetime = '28/01/2017'; -- Any date from the target month/year

select MyDate -- raw date or ...
    -- convert(varchar, MyDate, 103) + ' - ' + dateName(dw, MyDate) -- as Sample
    as WeekEndDate
from (
    select dateAdd(dd, number, dateAdd(mm, dateDiff(mm, 0, @OneDate), 0)) as MyDate
    from master..spt_values
    where type = 'P' and number < 31
    ) j
where 1 + (datePart(dw, MyDate) + @@DATEFIRST + 5) % 7 in (6, 7)
    and month(MyDate) = month(@OneDate)
-- order by 1 + (datePart(dw, MyDate) + @@DATEFIRST + 5) % 7, MyDate -- as Sample
;

Answer 6

解决此问题的另一种方法如下-

DECLARE @MONTH INT,@YEAR INT
SET @MONTH = 1;
SET @YEAR = 2017;

Declare @StartDate date =CAST(CAST(@MONTH AS VARCHAR(2)) + '/01/' + CAST(@YEAR AS VARCHAR(4)) AS [Date]), @EndDate date
Set @EndDate = EOMONTH(@StartDate)

Declare @Temp table (DateOfDay date, DaysName varchar(50))
While(@StartDate <= @EndDate)
Begin

    Insert into @Temp
    SELECT  @StartDate DateOfMonth, 
            case when DATENAME(DW, @StartDate) = 'Saturday' then DATENAME(DW, @StartDate) 
                 when DATENAME(DW, @StartDate) = 'Sunday' then DATENAME(DW, @StartDate) 
            end DaysName

    set @StartDate = DATEADD(d,1,@StartDate)
End

select * from @Temp where DaysName is not null order by DaysName, DateOfDay

Answer 7

你不能这样做吗？

SELECT DATENAME(dw,'10/11/2016') AS DATE 
WHERE DATE CONTAINS('Saturday') OR DATE CONTAINS('SUNDAY')

而不是“2016 年 10 月 11 日”，您只需要弄清楚如何生成月/年的所有日期？