如何使用DATEDIFF返回年、月、日?

以下是我对 Eric 功能的解决方案：

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'
)

如果开始日期在结束日期之后，则使用 ABS 进行良好的调用。

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

..或用于 SQL Server 2000 及之前的非 CTE：

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