How to use DATEDIFF to return year, month and day?

Question

How can I use DATEDIFF to return the difference between two dates in years, months and days in SQL Server 2005

DATEDIFF (date , dat         


        

Answer 1

Check this page... http://www.sqlteam.com/article/datediff-function-demystified

Create this functions:

CREATE FUNCTION dbo.fnYearsApart
(
        @FromDate DATETIME,
        @ToDate DATETIME
)
RETURNS INT
AS
BEGIN
        RETURN  CASE
                       WHEN @FromDate > @ToDate THEN NULL
                       WHEN DATEPART(day, @FromDate) > DATEPART(day, @ToDate) THEN DATEDIFF(month, @FromDate, @ToDate) - 1
                       ELSE DATEDIFF(month, @FromDate, @ToDate)
               END / 12
END

and

CREATE FUNCTION dbo.fnMonthsApart
(
        @FromDate DATETIME,
        @ToDate DATETIME
)
RETURNS INT
AS
BEGIN
        RETURN  CASE
                       WHEN @FromDate > @ToDate THEN NULL
                       WHEN DATEPART(day, @FromDate) > DATEPART(day, @ToDate) THEN DATEDIFF(month, @FromDate, @ToDate) - 1
                       ELSE DATEDIFF(month, @FromDate, @ToDate)
               END
END

and finally

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

    SET @yy = dbo.fnYearsApart(@dstart, @dend)  --DATEDIFF(yy, @dstart, @dend)
    SET @mm = dbo.fnMonthsApart(@dstart, @dend) --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) + ' años, ' + Convert(varchar(10),@getmm) + ' meses, '  + Convert(varchar(10),@getdd) + ' días'
    )
END

Greats!