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

CREATE FUNCTION FindDateDiff(@Date1 date,@Date2 date)
RETURNS TABLE 
AS
RETURN 
(
    SELECT
        CALC.Years,CALC.Months,D.Days,
        Result = RTRIM(Case When CALC.Years > 0 Then CONCAT(CALC.Years, ' year(s) ') Else '' End
                     + Case When CALC.Months > 0 Then CONCAT(CALC.Months, ' month(s) ') Else '' End
                     + Case When D.Days > 0 OR (CALC.Years=0 AND CALC.Months=0) Then CONCAT(D.Days, ' day(s)') Else '' End)
    FROM (VALUES(IIF(@Date1<@Date2,@Date1,@Date2),IIF(@Date1<@Date2,@Date2,@Date1))) T(StartDate, EndDate)
    CROSS APPLY(Select
        TempEndYear = Case When ISDATE(CONCAT(YEAR(T.EndDate), FORMAT(T.StartDate,'-MM-dd')))=1 Then CONCAT(YEAR(T.EndDate), FORMAT(T.StartDate,'-MM-dd'))
                        Else CONCAT(YEAR(T.EndDate),'-02-28') End
    ) TEY
    CROSS APPLY(Select EndYear = Case When TEY.TempEndYear > T.EndDate Then DATEADD(YEAR, -1, TEY.TempEndYear) Else TEY.TempEndYear End) EY
    CROSS APPLY(Select
        Years = DATEDIFF(YEAR,T.StartDate,EY.EndYear),
        Months = DATEDIFF(MONTH,EY.EndYear,T.EndDate)-IIF(DAY(EY.EndYear)>DAY(T.EndDate),1,0)
    ) CALC
    CROSS APPLY(Select Days =  DATEDIFF(DAY,DATEADD(MONTH,CALC.Months,DATEADD(YEAR,CALC.Years,T.StartDate)),T.EndDate)) D
)

and

Select * From dbo.FindDateDiff('2015-12-15','2018-12-14')

Result: