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

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

Answer 2

-- Swift 4 --

     let num = 1
     let formatter = NumberFormatter()
     formatter.numberStyle = .ordinal
     let day = formatter.string(from: NSNumber(value: num))

     print(day!)
     result - 1st

Answer 3

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...

Answer 4

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

Answer 5

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"
        }
    }

}

Answer 6

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