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.

+(NSString*)ordinalNumberFormat:(NSInteger)num{
    NSString *ending;

    int ones = num % 10;
    int tens = floor(num / 10);
    tens = tens % 10;
    if(tens == 1){
        ending = @"th";
    }else {
        switch (ones) {
            case 1:
                ending = @"st";
                break;
            case 2:
                ending = @"nd";
                break;
            case 3:
                ending = @"rd";
                break;
            default:
                ending = @"th";
                break;
        }
    }
    return [NSString stringWithFormat:@"%d%@", num, ending];
}

Adapted from nickf's answer here Is there an easy way in .NET to get "st", "nd", "rd" and "th" endings for numbers?

This was really useful. I created a gist with this code as an NSString category: https://gist.github.com/3119444 — dreadpirateryan, Jul 16 '12 at 00:41
Is the `floor` call needed? Casting `num / 10` to an `int` should remove anything after the decimal. Right? — Dylan Bettermann, Jan 31 '15 at 22:17
This is now possible with `numberFormatter.numberStyle = NSNumberFormatterOrdinalStyle` in iOS 9: http://stackoverflow.com/a/37106084/746890 — Chris Nolet, May 09 '16 at 00:14

Chris Nolet · Answer 1 · 2018-05-11T20:55:28.410

44

The correct way to do this from iOS 9 onwards, is:

NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
numberFormatter.numberStyle = NSNumberFormatterOrdinalStyle;

NSLog(@"%@", [numberFormatter stringFromNumber:@(1)]); // 1st
NSLog(@"%@", [numberFormatter stringFromNumber:@(2)]); // 2nd
NSLog(@"%@", [numberFormatter stringFromNumber:@(3)]); // 3rd, etc.

Alternatively:

NSLog(@"%@", [NSString localizedStringFromNumber:@(1)
                                     numberStyle:NSNumberFormatterOrdinalStyle]); // 1st

edited May 11 '18 at 20:55

answered May 08 '16 at 23:54

Chris Nolet

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10

There is also `NumberFormatter.localizedString(from: n, number: .ordinal)` already from (I think) iOS 4 onwards. – Drux Oct 03 '16 at 19:59
That method (`localizedStringFromNumber:numberStyle:` in Objective-C) has been around for a while, but they only added `.ordinal` (or `NSNumberFormatterOrdinalStyle`) in iOS9 unfortunately. You can use that form if you're targetting iOS 9 onwards, though. – Chris Nolet Oct 03 '16 at 22:35

Magoo · Answer 2 · 2018-12-11T03:56:16.353

26

As of iOS 9

Swift 4

private var ordinalFormatter: NumberFormatter = {
    let formatter = NumberFormatter()
    formatter.numberStyle = .ordinal
    return formatter
}()

extension Int {
    var ordinal: String? {
        return ordinalFormatter.string(from: NSNumber(value: self))
    }
}

It's probably best to have the formatter outside the extension...

edited Dec 11 '18 at 03:56

answered Feb 23 '17 at 17:00

Magoo

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Swift 3 version would be... let formatter = NumberFormatter() formatter.numberStyle = .ordinal formatter.string(from: NSNumber(value: value)) – Dimitris Mar 11 '17 at 13:43
Cheers.. though you could just edit... and maybe have just written the correction rather than the entire thing. – Magoo Mar 13 '17 at 09:24

score 22 · Answer 3 · edited May 23 '17 at 10:31

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;
}

score 18 · Answer 4 · edited Nov 12 '16 at 01:49

Other Swift solutions do not produce correct result and contain mistakes. I have translated CmKndy solution to Swift

extension Int {

    var ordinal: String {
        var suffix: String
        let ones: Int = self % 10
        let tens: Int = (self/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"
        }
        return "\(self)\(suffix)"
    }

}

test result: 0th 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th 11th 12th 13th 14th 15th 16th 17th 18th 19th 20th 21st 22nd 23rd

Abizern · Accepted Answer · 2010-12-23T19:48:10.050

Since the question asked for a number formatter, here's a rough one I made.

//
//  OrdinalNumberFormatter.h
//

#import <Foundation/Foundation.h>


@interface OrdinalNumberFormatter : NSNumberFormatter {

}

@end

and the implementation:

//
//  OrdinalNumberFormatter.m
//

#import "OrdinalNumberFormatter.h"


@implementation OrdinalNumberFormatter

- (BOOL)getObjectValue:(id *)anObject forString:(NSString *)string errorDescription:(NSString **)error {
    NSInteger integerNumber;
    NSScanner *scanner;
    BOOL isSuccessful = NO;
    NSCharacterSet *letters = [NSCharacterSet letterCharacterSet];

    scanner = [NSScanner scannerWithString:string];
    [scanner setCaseSensitive:NO];
    [scanner setCharactersToBeSkipped:letters];

    if ([scanner scanInteger:&integerNumber]){
        isSuccessful = YES;
        if (anObject) {
            *anObject = [NSNumber numberWithInteger:integerNumber];
        }
    } else {
        if (error) {
            *error = [NSString stringWithFormat:@"Unable to create number from %@", string];
        }
    }

    return isSuccessful;
}

- (NSString *)stringForObjectValue:(id)anObject {
    if (![anObject isKindOfClass:[NSNumber class]]) {
        return nil;
    }

    NSString *strRep = [anObject stringValue];
    NSString *lastDigit = [strRep substringFromIndex:([strRep length]-1)];

    NSString *ordinal;


    if ([strRep isEqualToString:@"11"] || [strRep isEqualToString:@"12"] || [strRep isEqualToString:@"13"]) {
        ordinal = @"th";
    } else if ([lastDigit isEqualToString:@"1"]) {
        ordinal = @"st";
    } else if ([lastDigit isEqualToString:@"2"]) {
        ordinal = @"nd";
    } else if ([lastDigit isEqualToString:@"3"]) {
        ordinal = @"rd";
    } else {
        ordinal = @"th";
    }

    return [NSString stringWithFormat:@"%@%@", strRep, ordinal];
}

@end

Instantiate this as an Interface Builder object and attach the Text Field's formatter outlet to it. For finer control (such as setting maximum and minimum values, you should create an instance of the formatter, set the properties as you wish and attach it to text field using it's setFormatter: method.

You can download the class from GitHub (including an example project)

@Abizem like jan pointed it this misses some of the teens however I see that it is fixed in your GitHub repo i.e. [strRepresentation hasSuffix:@"11"] — Steve Moser, Jul 16 '13 at 02:44
Thanks @Abizern! Gist with Abizern's code translated into Swift: https://gist.github.com/yoiang/bad782fb55a3cb8d062e — yo.ian.g, Nov 13 '15 at 04:25

Krunal Patel · Answer 6 · 2021-06-01T08:46:52.510

14

-- Swift 4/5 --

     let num = 1
     let formatter = NumberFormatter()
     formatter.numberStyle = .ordinal
     let day = formatter.string(from: NSNumber(value: num))
     
     print(day!)
     result - 1st

edited Jun 01 '21 at 08:46

answered Dec 29 '17 at 10:11

Krunal Patel

1,649
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superarts.org · Answer 7 · 2015-08-28T04:09:47.393

7

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

edited Aug 28 '15 at 04:09

answered Nov 14 '14 at 06:17

superarts.org

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This would create 13rd instead of 13th, 12nd instead 12th. – Swinny89 Aug 27 '15 at 12:22
Thanks @Swinny89 - it's fixed. – superarts.org Aug 28 '15 at 04:10

score 5 · Answer 8 · answered Jun 07 '12 at 00:39

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;
}

score 5 · Answer 9 · answered Mar 15 '16 at 19:46

5

Here's a compact Swift extension suitable for all integer types:

extension IntegerType {
    func ordinalString() -> String {
        switch self % 10 {
        case 1...3 where 11...13 ~= self % 100: return "\(self)" + "th"
        case 1:    return "\(self)" + "st"
        case 2:    return "\(self)" + "nd"
        case 3:    return "\(self)" + "rd"
        default:   return "\(self)" + "th"
        }
    }
}

Example usage:

let numbers = (0...30).map { $0.ordinalString() }
print(numbers.joinWithSeparator(", "))

Output:

0th, 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th, 11th, 12th, 13th, 14th, 15th, 16th, 17th, 18th, 19th, 20th, 21st, 22nd, 23rd, 24th, 25th, 26th, 27th, 28th, 29th, 30th

answered Mar 15 '16 at 19:46

maxkonovalov

3,651
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~= what does this operator do? – ManicMonkOnMac Mar 15 '16 at 19:56
2

~= is a built-in pattern match operator, in this case it checks if `self % 100` falls in the range `[11...13]`, to handle the "tenths" case correctly. For more info see: https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Patterns.html – maxkonovalov Mar 15 '16 at 20:00
It's been there since swift 1.x I believe ;) – maxkonovalov Mar 15 '16 at 20:03
I meant the link you shared, it words the whole patterns thing in a new way which wasn't there before – ManicMonkOnMac Mar 15 '16 at 20:06

ymutlu · Answer 10 · 2020-01-24T08:25:17.347

5

There is a simple solution for this

Swift

let formatter = NumberFormatter()
formatter.numberStyle = .ordinal
let first = formatter.string(from: 1) // 1st
let second = formatter.string(from: 2) // 2nd

Obj-c

NSNumberFormatter *numberFormatter = [NSNumberFormatter new];
numberFormatter.numberStyle = NSNumberFormatterOrdinalStyle;
NSString* first = [numberFormatter stringFromNumber:@(1)]; // 1st
NSString* second = [numberFormatter stringFromNumber:@(2)]; // 2nd

Referance: hackingwithswift.com

edited Jan 24 '20 at 08:25

answered Aug 28 '17 at 09:20

ymutlu

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is there a way to get it as first / second , third etc – Nassif Apr 02 '20 at 09:53

Dave DeLong · Answer 11 · 2010-07-22T20:44:53.357

3

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.

edited Jul 22 '10 at 20:44

answered Jul 22 '10 at 20:11

Dave DeLong

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This misses some of the teens. I found the algorithm here: http://stackoverflow.com/questions/69262/is-there-an-easy-way-in-net-to-get-st-nd-rd-and-th-endings-for-numbers Just thought maybe the formatter could handle. – jan Jul 22 '10 at 20:16
@jan good point about 11, 12, and 13. Those suffixes could easily be special-cased. – Dave DeLong Jul 22 '10 at 20:28

score 3 · Answer 12 · answered Feb 21 '16 at 20:41

A clean Swift version (for English only):

func ordinal(number: Int) -> String {
    if (11...13).contains(number % 100) {
        return "\(number)th"
    }
    switch number % 10 {
        case 1: return "\(number)st"
        case 2: return "\(number)nd"
        case 3: return "\(number)rd"
        default: return "\(number)th"
    }
}

Can be done as an extension for Int:

extension Int {

    func ordinal() -> String {
        return "\(self)\(ordinalSuffix())"
    }

    func ordinalSuffix() -> String {
        if (11...13).contains(self % 100) {
            return "th"
        }
        switch self % 10 {
            case 1: return "st"
            case 2: return "nd"
            case 3: return "rd"
            default: return "th"
        }
    }

}

score 2 · Answer 13 · answered Jul 22 '10 at 20:41

2

The following example demonstrates how to handle any number. It's in c# however it can easily converted to any language.

http://www.bytechaser.com/en/functions/b6yhfyxh78/convert-number-to-ordinal-like-1st-2nd-in-c-sharp.aspx

answered Jul 22 '10 at 20:41

Alex

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score 2 · Answer 14 · edited Dec 28 '15 at 12:13

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;
}

score 1 · Answer 15 · edited Feb 26 '20 at 20:44

Here's a Swift solution that cycles through the user's preferred languages until it finds one with known rules (which are pretty easy to add) for ordinal numbers:

    extension Int {
        var localizedOrdinal: String {

            func ordinalSuffix(int: Int) -> String {
                for language in NSLocale.preferredLanguages() as [String] {

                switch language {
                    case let l where l.hasPrefix("it"):
                        return "°"
                    case let l where l.hasPrefix("en"):
                        switch int {
                        case let x where x != 11 && x % 10 == 1:
                            return "st"
                        case let x where x != 12 && x % 10 == 2:
                            return "nd"
                        case let x where x != 13 && x % 10 == 3:
                            return "rd"
                        default:
                            return "th"
                        }
                    default:
                        break
                    }
                }

                return ""
            }

            return "\(self)" + ordinalSuffix(self)
        }
    }

score 0 · Answer 16 · answered Aug 19 '15 at 21:24

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);
}

score 0 · Answer 17 · answered Sep 13 '16 at 19:12

Here's a short Int extension for the English language that also accounts for and displays negative integers correctly:

extension Int {
    func ordinal() -> String {
        let suffix: String!
        // treat negative numbers as positive for suffix
        let number = (self < 0 ? self * -1 : self)

        switch number % 10 {
        case 0:
            suffix = self != 0 ? "th" : ""
        case 1:
            suffix = "st"
        case 2:
            suffix = "nd"
        case 3:
            suffix = "rd"
        default:
            suffix = "th"
        }

        return String(self) + suffix
    }
}

score 0 · Answer 18 · answered May 05 '17 at 07:04

- (NSString *) formatOrdinalNumber:(NSInteger )number{
    NSString *result = nil;
    //0 remains just 0
    if (number == 0) {
        result = @"0";
    }

    //test for number between 3 and 21 as they follow a
    //slightly different rule and all end with th
    else if (number > 3 && number < 21)
    {
        result = [NSString stringWithFormat:@"%ld th",(long)number];
    }
    else {
        //return the last digit of the number e.g. 102 is 2
        NSInteger lastdigit = number % 10;
        switch (lastdigit)
        {
            case 1: result = [NSString stringWithFormat:@"%ld st",(long)number]; break;
            case 2: result = [NSString stringWithFormat:@"%ld nd",(long)number]; break;
            case 3: result = [NSString stringWithFormat:@"%ld rd",(long)number]; break;
            default: result = [NSString stringWithFormat:@"%ld th",(long)number];
        }
    }
    return result;
}

score 0 · Answer 19 · edited Apr 19 '19 at 13:16

0

You can try this, Its well simplified.

function numberToOrdinal(n) {

  if (n==0) {
    return n;
   }
   var j = n % 10,
       k = n % 100;


   if (j == 1 && k != 11) {
       return n + "st";
   }
   if (j == 2 && k != 12) {
       return n + "nd";
   }
   if (j == 3 && k != 13) {
       return n + "rd";
   }
   return n + "th";
}

edited Apr 19 '19 at 13:16

J. Doe

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answered Apr 19 '19 at 12:44

Folalu Timothy

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score 0 · Answer 20 · answered Sep 15 '22 at 06:57

public extension Int {
 var ordinalValue: String? {
    let formatter = NumberFormatter()
    formatter.numberStyle = .ordinal
    formatter.locale = Locale(identifier: "en_US")
    return formatter.string(from: NSNumber(value: self))
 }
}

Don't forget to add the Locale, if it is not added, it won't work. Usage:

let number = 2
let ordinalNumber = number.ordinalValue
print(ordinalNumber) //It will print 2nd

score -4 · Answer 21 · answered Feb 03 '14 at 15:51

This was my brute force implementation to taking a NSString* representation of the date and returning the ordinal value. I feel it's much easier to read.

NSDictionary *ordinalDates = @{
    @"1": @"1st",
    @"2": @"2nd",
    @"3": @"3rd",
    @"4": @"4th",
    @"5": @"5th",
    @"6": @"6th",
    @"7": @"7th",
    @"8": @"8th",
    @"9": @"9th",
    @"10": @"10th",
    @"11": @"11th",
    @"12": @"12th",
    @"13": @"13th",
    @"14": @"14th",
    @"15": @"15th",
    @"16": @"16th",
    @"17": @"17th",
    @"18": @"18th",
    @"19": @"19th",
    @"20": @"20th",
    @"21": @"21st",
    @"22": @"22nd",
    @"23": @"23rd",
    @"24": @"24th",
    @"25": @"25th",
    @"26": @"26th",
    @"27": @"27th",
    @"28": @"28th",
    @"29": @"29th",
    @"30": @"30th",
    @"31": @"31st" };

How long does the list go? – Eneko Alonso Feb 21 '16 at 20:29 — Eneko Alonso, Feb 21 '16 at 20:29

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

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