Question

我遇到了问题。我想在一个月内使用函数得到星期日，虽然我已经使用一个过程获得了类似的结果，但我需要在我的select查询中调用该函数来返回每个select的结果。我给出了我的程序代码

    create proc [dbo].[getSundaysandSaturdays]
(
    @Year int=2016,
    @Month int=11,
    @fdays as int output
)
as
begin
    ;with dates as
    (
        select dateadd(month,@month-1,dateadd(year,@year-1900,0)) as StartDate
        union all
        select startdate + 1 from dates where month(startdate+1) = @Month
    )
    select @fdays=count(*) from dates where datediff(dd,0,startdate)%7 in (4)
    return @fdays
end

Answer 1

您当前的问题是您无法从选择中调用存储过程。你需要一个功能。这是一个有效的功能。

另一个问题 - 避免让递归的sql计算一个月内的星期日。这个“简单”的表达式将计算一个月内的星期日。

CREATE function f_count_sundays
(
  @Year int,
  @Month int
) returns int
BEGIN
DECLARE @x date = dateadd(month,@month-1+(@year-1900)*12,6)
RETURN datediff(d, dateadd(d,datediff(d,-1,@x)/7*7,-1),dateadd(m,1,@x))/7
END

go

SELECT dbo.f_count_sundays(2016,10)

结果：

Answer 2

我已经改变如下，并修改了使用datepart函数查找星期日所需的逻辑。如果你想返回星期六的计数，那么将它添加到where子句中。

 create function [dbo].[getSundaysandSaturdays]
    (
        @Year int,
        @Month int

    ) 
    RETURNS int
    as 
    begin
    declare     @fdays int 
        ;with dates as
        (
            select dateadd(month,@month-1,dateadd(year,@year-1900,0)) as StartDate
            union all
            select startdate + 1 from dates where month(startdate+1) = @Month
        )
        select @fdays=count(*) from dates where datepart(dw,StartDate) =1
      return @fdays
      end

和呼叫功能如下

  select  dbo.[getSundaysandSaturdays](2015,11)

Answer 3

你试过这个吗？

select * from f_table_date('20090601', '20090630')
where weekdayname ='sun'

Answer 4

您是否考虑过创建一个Date表然后再查询，因为这可能是一个比您尝试做的更好的长期解决方案，因为您可以计算一个月中的任何一天，比SQL函数更快，更有说服力

SELECT Count(0) 
FROM dimDate 
WHERE 
DayName = 'Sunday'
AND [CalendarYear] = 2016
AND [CalendarMonth] = 'May' 
-- Or you could use a '03' if you preferred.

Answer 5

CREATE FUNCTION [DBO].[UDF_getSundaysandSaturdays] 
(
    @Year int,
    @Month int,
    @Day int  --5 for Sat & 6 for Sun
    )
RETURNS INT
as
begin
DECLARE @DayCount INT
    ;with dates as
    (
        select dateadd(month,@month-1,dateadd(year,@year-1900,0)) as StartDate
        union all
        select startdate + 1 from dates where month(startdate+1) = @Month
    )

    SELECT @DayCount=(select count(*) from dates where datediff(dd,0,startdate)%7 in (@Day))
    RETURN @DayCount
end

在函数SQL中获取给定月份的星期日

5 个答案: