How do you calculate the day of the year for a specific date in Objective C

Question

This is something I found myself spending hours to figure out and therefor wants to share with you.

So the question was, How do i determine the day of the year for a specific date?

e.g. January 15 is the 15'th day and December 31 is the 365'th day when it's not leap year.

Answer 1

With the NSDate class. Use message timeIntervalSinceDate. It will return you a NSTimeInterval value (actually it's a double) that represents the seconds elapsed since the date you want. After that, it's easy to convert seconds-to-days.

If you truncate seconds/86400 will give you days.

Let's say you want days since January 1st 2010.

// current date/time
NSDate *now = [[NSData alloc] init]; 

// seconds elapsed since January 1st 2010 00:00:00 (GMT -4) until now
NSInteval interval = [now timeIntervalSinceDate: [NSDate dateWithString:@"2010-01-01 00:00:00 -0400"]];

// days since January 1st 2010 00:00:00 (GMT -4) until now
int days = (int)interval/86400;

Answer 2

So the solution I came up with is pretty neat and simple and not in any way complicated.

NSDateFormatter *formatter = [[NSDateFormatter alloc] init];
[formatter setDateFormat:@"D"];
NSUInteger dayOfYear = [[formatter stringFromDate:[NSDate date]] intValue];
[formatter release];
return dayOfYear;

The trick here which I spent so long time to figure out was to use the NSDateFormatter. The "D" is the flag for day of year.

Hope that this was helpful to you who have the same problem as I had.

Answer 3

Try this:

NSCalendar *gregorian =
   [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSUInteger dayOfYear =
   [gregorian ordinalityOfUnit:NSDayCalendarUnit
     inUnit:NSYearCalendarUnit forDate:[NSDate date]];
[gregorian release];
return dayOfYear;

where date is the date you want to determine the day of the year for. The documentation on the NSCalendar.ordinalityOfUnit method is here.