Math with datetime??

TheMaj0r

Hello, I have a datetime the holds a birthdate, for example 5/1/1984, I want to create a method that takes this date and return a string for the age ,for example 23 years and 9 months and 30 days

Thanks

BinaryCoder

These "age" problems are more complex than they appear on the surface.

To just handle the "years" part, consider this method:

Code Block

// Returns the number of complete years (age) between d1 and d2.

public static int CalculateAge(DateTime d1, DateTime d2)

{

int years = d2.Year - d1.Year;

DateTime anniversary = new DateTime(d2.Year, d1.Month, d1.Day) + d1.TimeOfDay;

if (anniversary > d2)

years -= 1; // Check if did not yet reach the anniversary DateTime.

return years;

}

By months and days, you need to define what "months and days" mean: not all months have the same number of days. You could easily adapt the above to find year and days:

int days = (d2 - anniversary).Days;

TheMaj0r

Thanks, The most complex is computing the months and days as we dont know the days of months and how many leap years were within the two dates

I quickly thought about this easy solution, but it is not 100% accurate:

DateTime dob = DateTime.Parse(TextBox1.Text);

int days = ((TimeSpan)(DateTime.Now - dob)).Days;

int years = (int)(days / 365.25);

int months = (int)(days / 30.4375);

int involvedMonths = years * 12;

int monthsSlack = months - involvedMonths;

int involvedDays = (int)(monthsSlack * 30.4375);

int daysSlack = (int)((days - (years * 365.25)) - involvedDays);

Label1.Text = years.ToString() + "Years, " + monthsSlack.ToString() + "Months, " + daysSlack.ToString() + "Days.";

BinaryCoder

There is no "one" answer to this problem. It is all a matter of how you define "age". You haven't provided a rock solid definition of "age" (which, of course, needs to say what to do about the differences in the number of days per month).

TheMaj0r

I just simplified it and defined average of days per month which is about 30.4

boban.s

Check my post here:
http://forums.microsoft.com/MSDN/ShowPost.aspx PostID=2257780&SiteID=1

TheMaj0r

Thanks worked accuretly