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t-sqldatediff

How to calculate age in T-SQL with years, months, and days


What would be the best way to calculate someone's age in years, months, and days in T-SQL (SQL Server 2000)?

The datediff function doesn't handle year boundaries well, plus getting the months and days separate will be a bear. I know I can do it on the client side relatively easily, but I'd like to have it done in my stored procedure.


Solution

  • Here is some T-SQL that gives you the number of years, months, and days since the day specified in @date. It takes into account the fact that DATEDIFF() computes the difference without considering what month or day it is (so the month diff between 8/31 and 9/1 is 1 month) and handles that with a case statement that decrements the result where appropriate.

    DECLARE @date datetime, @tmpdate datetime, @years int, @months int, @days int
    SELECT @date = '2/29/04'
    
    SELECT @tmpdate = @date
    
    SELECT @years = DATEDIFF(yy, @tmpdate, GETDATE()) - CASE WHEN (MONTH(@date) > MONTH(GETDATE())) OR (MONTH(@date) = MONTH(GETDATE()) AND DAY(@date) > DAY(GETDATE())) THEN 1 ELSE 0 END
    SELECT @tmpdate = DATEADD(yy, @years, @tmpdate)
    SELECT @months = DATEDIFF(m, @tmpdate, GETDATE()) - CASE WHEN DAY(@date) > DAY(GETDATE()) THEN 1 ELSE 0 END
    SELECT @tmpdate = DATEADD(m, @months, @tmpdate)
    SELECT @days = DATEDIFF(d, @tmpdate, GETDATE())
    
    SELECT @years, @months, @days