NSDateFormatter dateFromString无

Question

I know the question exists already, but the given answers don't help me out.

birthday has the value: Jun 29, 2013, 8:45 AM

Here is my code:

NSDate* now = [[NSDate alloc] init];
NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setTimeStyle:NSDateFormatterShortStyle];
[dateFormatter setDateStyle:NSDateFormatterMediumStyle];
[dateFormatter setLocale:[NSLocale currentLocale]];
[dateFormatter setFormatterBehavior:NSDateFormatterBehaviorDefault];
NSDate *birthdayDate = [dateFormatter dateFromString:birthday];

NSLog(@"%@", birthdayDate);

birthdayDate is always nil

What should I do?

Answer 1

If you're working with fixed-format dates, you should first set the locale of the date formatter to something appropriate for your fixed format. In most cases the best locale to choose is en_US_POSIX, a locale that's specifically designed to yield US English results regardless of both user and system preferences. en_US_POSIX is also invariant in time (if the US, at some point in the future, changes the way it formats dates, en_US will change to reflect the new behavior, but en_US_POSIX will not), and between platforms (en_US_POSIX works the same on iPhone OS as it does on OS X, and as it does on other platforms).

// String to Date
NSString *birthday = @"Jun 29, 2013, 8:45 AM";

NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat:@"MMM dd, yyyy, HH:mm a"];
dateFormatter.timeZone = [NSTimeZone timeZoneForSecondsFromGMT:0];
// Setting the locale to POSIX ensures that
// the user's locale won't be used
dateFormatter.locale = [[NSLocale alloc] initWithLocaleIdentifier:@"en_US_POSIX"];
NSDate *birthdayDate = [dateFormatter dateFromString:birthday];

NSLog(@"birthdayDate:%@", birthdayDate);

// Date to String
[dateFormatter setLocale:[NSLocale currentLocale]];
dateFormatter.timeZone = [NSTimeZone systemTimeZone];
NSString *originalBirthday = [dateFormatter stringFromDate:birthdayDate];

NSLog(@"originalBirthday=%@",originalBirthday);