NSNumberFormatter and 'th' 'st' 'nd' 'rd' (ordinal) number endings

Question

Is there a way to use NSNumberFormatter to get the \'th\' \'st\' \'nd\' \'rd\' number endings?

EDIT:

Looks like it does not exist. Here\'s what I\'m using.

Answer 1

This will convert date to string and also add ordinal in the date. You can modify the date formatte by changing NSDateFormatter object

-(NSString*) getOrdinalDateString:(NSDate*)date
{
    NSString* string=@"";
    NSDateComponents *components = [[NSCalendar currentCalendar] components: NSCalendarUnitDay fromDate:date];

    if(components.day == 1 || components.day == 21 || components.day == 31)
        string = @"st";

    else if (components.day == 2 || components.day == 22)
        string = @"nd";

    else if (components.day == 3 || components.day == 23)
        string = @"rd";

    else
        string = @"th";


    NSDateFormatter *dateFormatte = [[NSDateFormatter alloc] init];
    [dateFormatte setFormatterBehavior:NSDateFormatterBehavior10_4];
    [dateFormatte setDateFormat:[NSString stringWithFormat:@"d'%@' MMMM yyyy",string]];

    NSString *dateString = [dateFormatte stringFromDate:date];
    return dateString;
}

Answer 2

I'm not aware of this capability. However, it's possible to do this yourself. In English, the ordinal (th, st, nd, rd, etc) has a really simple pattern:

If the number ends with: => Use:

• 0 => th
• 1 => st
• 2 => nd
• 3 => rd
• 4 => th
• 5 => th
• 6 => th
• 7 => th
• 8 => th
• 9 => th
• 11 => th
• 12 => th
• 13 => th

This will not spell out the word for you, but it will allow you to do something like: "42nd", "1,340,697th", etc.

This gets more complicated if you need it localized.

Answer 3

This does the trick in one method (for English). Thanks nickf https://stackoverflow.com/a/69284/1208690 for original code in PHP, I just adapted it to objective C:-

-(NSString *) addSuffixToNumber:(int) number
{
    NSString *suffix;
    int ones = number % 10;
    int tens = (number/10) % 10;

    if (tens ==1) {
        suffix = @"th";
    } else if (ones ==1){
        suffix = @"st";
    } else if (ones ==2){
        suffix = @"nd";
    } else if (ones ==3){
        suffix = @"rd";
    } else {
        suffix = @"th";
    }

    NSString * completeAsString = [NSString stringWithFormat:@"%d%@", number, suffix];
    return completeAsString;
}

Answer 4

Many of the solutions here don't handle higher numbers like 112. Here is a simple way to do it.

for(int i=0;i<1000;i++){
    int n = i;
    NSString* ordinal = @"th";
    if(n%10==1 && n%100!=11) ordinal = @"st";
    if(n%10==2 && n%100!=12) ordinal = @"nd";
    if(n%10==3 && n%100!=13) ordinal = @"rd";
    NSLog(@"You are the %d%@",i,ordinal);
}

Answer 5

It's quite simple in English. Here's a swift extension:

extension Int {
    var ordinal: String {
        get {
            var suffix = "th"
            switch self % 10 {
                case 1:
                    suffix = "st"
                case 2:
                    suffix = "nd"
                case 3:
                    suffix = "rd"
                default: ()
            }
            if 10 < (self % 100) && (self % 100) < 20 {
                suffix = "th"
            }
            return String(self) + suffix
        }
    }
}

Then call something like:

    cell.label_position.text = (path.row + 1).ordinal

Answer 6

Just adding another implementation as a class method. I didn't see this question posted until after I implemented this from an example in php.

+ (NSString *)buildRankString:(NSNumber *)rank
{
    NSString *suffix = nil;
    int rankInt = [rank intValue];
    int ones = rankInt % 10;
    int tens = floor(rankInt / 10);
    tens = tens % 10;
    if (tens == 1) {
        suffix = @"th";
    } else {
        switch (ones) {
            case 1 : suffix = @"st"; break;
            case 2 : suffix = @"nd"; break;
            case 3 : suffix = @"rd"; break;
            default : suffix = @"th";
        }
    }
    NSString *rankString = [NSString stringWithFormat:@"%@%@", rank, suffix];
    return rankString;
}