如何使用DATEDIFF返回年，月和日？

Question

如何使用DATEDIFF返回年份，月份和天中两个日期之间的差SQL Server 2005

DATEDIFF (date , date)

结果如何：2年3个月10天

有人能完成t-sql吗？

ALTER FUNCTION [dbo].[gatYMD](@dstart VARCHAR(50), @dend VARCHAR(50))
RETURNS VARCHAR(50) AS
BEGIN
    DECLARE @yy INT
    DECLARE @mm INT
    DECLARE @getmm INT
    DECLARE @dd INT

    SET @yy = DATEDIFF(yy, @dstart, @dend)
    SET @mm = DATEDIFF(mm, @dstart, @dend)
    SET @dd = DATEDIFF(dd, @dstart, @dend)
    SET @getmm = ABS(DATEDIFF(mm, DATEADD(yy, @yy, @dstart), @dend))

    RETURN (
        Convert(varchar(10),@yy) + 'year' + Convert(varchar(10),@mm) + 'month'  + Convert(varchar(10),@dd) + 'day'
        )
END

Answer 1

这是我对Eric函数的解决方案：

DECLARE @getmm INT
DECLARE @getdd INT

SET @yy = DATEDIFF(yy, @dstart, @dend)
SET @mm = DATEDIFF(mm, @dstart, @dend)
SET @dd = DATEDIFF(dd, @dstart, @dend)
SET @getmm = ABS(DATEDIFF(mm, DATEADD(yy, @yy, @dstart), @dend))
SET @getdd = ABS(DATEDIFF(dd, DATEADD(mm, DATEDIFF(mm, DATEADD(yy, @yy, @dstart), @dend), DATEADD(yy, @yy, @dstart)), @dend))

RETURN (
  Convert(varchar(10),@yy) + 'year' + Convert(varchar(10),@getmm) + 'month'  + Convert(varchar(10),@getdd) + 'day'
)

如果开始日期晚于结束日期，请使用ABS进行处理。

这个：

WITH ex_table AS (
  SELECT '2007-01-01' 'birthdatetime',
         '2009-03-29' 'visitdatetime')
SELECT CAST(DATEDIFF(yy, t.birthdatetime, t.visitdatetime) AS varchar(4)) +' year '+
       CAST(DATEDIFF(mm, DATEADD(yy, DATEDIFF(yy, t.birthdatetime, t.visitdatetime), t.birthdatetime), t.visitdatetime) AS varchar(2)) +' month '+
       CAST(DATEDIFF(dd, DATEADD(mm, DATEDIFF(mm, DATEADD(yy, DATEDIFF(yy, t.birthdatetime, t.visitdatetime), t.birthdatetime), t.visitdatetime), DATEADD(yy, DATEDIFF(yy, t.birthdatetime, t.visitdatetime), t.birthdatetime)), t.visitdatetime) AS varchar(2)) +' day' AS result
  FROM ex_table t

..或非CTE用于SQL Server 2000及更低版本：

SELECT CAST(DATEDIFF(yy, t.birthdatetime, t.visitdatetime) AS varchar(4)) +' year '+
       CAST(DATEDIFF(mm, DATEADD(yy, DATEDIFF(yy, t.birthdatetime, t.visitdatetime), t.birthdatetime), t.visitdatetime) AS varchar(2)) +' month '+
       CAST(DATEDIFF(dd, DATEADD(mm, DATEDIFF(mm, DATEADD(yy, DATEDIFF(yy, t.birthdatetime, t.visitdatetime), t.birthdatetime), t.visitdatetime), DATEADD(yy, DATEDIFF(yy, t.birthdatetime, t.visitdatetime), t.birthdatetime)), t.visitdatetime) AS varchar(2)) +' day' AS result
  FROM (SELECT '2007-01-01' 'birthdatetime',
         '2009-03-29' 'visitdatetime') t

...将返回：

result
----------------------
2 year 2 month 28 day

参考：DATEDIFF

Answer 2

创建此函数，它将给出精确的日期差异，如年月日

    Create function get_Exact_Date_diff(@date smalldatetime,@date2 smalldatetime)
 returns varchar(50)

    as

    begin

    declare @date3 smalldatetime

    Declare @month int,@year int,@day int

     if @date>@date2
     begin
     set @date3=@date2
     set @date2=@date
     set @date=@date3
     end



    SELECT @month=datediff (MONTH,@date,@date2)

    if dateadd(month,@month,@date) >@date2
    begin
    set @month=@month-1
    end
    set @day=DATEDIFF(day,dateadd(month,@month,@date),@date2)

    set @year=@month/12
    set @month=@month % 12

    return (case when @year=0 then '' when @year=1 then convert(varchar(50),@year ) + ' year ' when @year>1 then convert(varchar(50),@year ) + ' years ' end)
    + (case when @month=0 then '' when @month=1 then convert(varchar(50),@month ) + ' month ' when @month>1 then convert(varchar(50),@month ) + ' months ' end)
    + (case when @day=0 then '' when @day=1 then convert(varchar(50),@day ) + ' day ' when @day>1 then convert(varchar(50),@day ) + ' days ' end)

    end