Check if a String is alphanumeric in Swift

Question

In Swift, how can I check if a String is alphanumeric, ie, if it contains only one or more alphanumeric characters [a-zA-Z0-9], excluding letters with diacritics, eg, é.

In Objective-C: http://stackoverflow.com/questions/1671605/how-to-check-if-a-string-only-contains-alphanumeric-characters-in-objective-c — ma11hew28, Mar 14 '16 at 16:23

ma11hew28 · Accepted Answer · 2018-06-18T13:23:38.417

81

extension String {
    var isAlphanumeric: Bool {
        return !isEmpty && range(of: "[^a-zA-Z0-9]", options: .regularExpression) == nil
    }
}

"".isAlphanumeric        // false
"abc".isAlphanumeric     // true
"123".isAlphanumeric     // true
"ABC123".isAlphanumeric  // true
"iOS 9".isAlphanumeric   // false

edited Jun 18 '18 at 13:23

answered Mar 14 '16 at 16:22

ma11hew28

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Why not make it a computed property instead? – Tim Vermeulen Mar 14 '16 at 17:02
4

You might want to use this regex, it's intended for this use: `^[:alnum:]+$` – Mar 14 '16 at 19:16
@TimVermeulen That'd work too. I guess I went with a method because it does quite a bit more checking than say `Set.isEmpty`. Also, [in Python, `str.isalnum()` is a method](https://docs.python.org/library/stdtypes.html#str.isalnum). – ma11hew28 Mar 14 '16 at 21:49
@KennethBruno I tried that, but in iOS 9.2, `[:alnum:]` includes letters with diacritics, e.g., é, which is not what I want. I just reported this issue to Apple. – ma11hew28 Mar 14 '16 at 23:29
Yes, it does. If you want to exclude letters outside the ASCII character set use: `^([:ascii:]|[:alnum])+$`. The benefit of this over the one in the answer is that your intent is clearer, it's plain you want only ASCII and alphanumeric characters. – Mar 15 '16 at 00:10
@MattDiPasquale I don't really agree with your reasoning, whether a string is alphanumeric or not is a property, no matter how it was computed. – Tim Vermeulen Mar 15 '16 at 12:44
1

@TimVermeulen well, another reason I initially made it a method is that Swift has methods like `UnicodeDecodingResult.isEmptyInput()` & `UnicodeScalar.isASCII()`. But, I just noticed that Swift also has properties like `String.isEmpty`, `StaticString.isASCII`, and `Float.isNaN`. So, I changed it to a property. I also made a couple Swift bug reports: https://bugs.swift.org/browse/SR-950 & https://bugs.swift.org/browse/SR-951. – ma11hew28 Mar 15 '16 at 18:47
Yeah, `UIKit` still isn't very consistent regarding properties and functions (among other things). Thanks for reporting, I voted for them. – Tim Vermeulen Mar 15 '16 at 18:52
@TimVermeulen sure. Thanks! :-) And, thank you for your feedback here. – ma11hew28 Mar 15 '16 at 19:16
@KennethBruno thank you, but `[a-zA-Z0-9]` is shorter & (to me) more explicit. I don't always remember (and so have to look up) which characters the `[:ascii:]` & `[:alnum:]` POSIX character classes include—Not to mention, acronyms (like ASCII) & abbreviations (like alnum) tend to sacrifice clarity. Also, I thought `(a|b)` meant `a` or `b`, so it's unclear to me that `([:ascii:]|[:alnum])` means `[:ascii:]` *and* `[:alnum]`. – ma11hew28 Mar 15 '16 at 19:32
The nature of the code is that we want to include anything that is `not (ascii or alphanumeric)`, translating that with [De Morgan's laws](https://en.wikipedia.org/wiki/De_Morgan%27s_laws) we get including anything that is `(not ascii) and (not alphanumeric)`. So you can express it with either `and` or `or`, it just takes knowing the algorithms deeply enough. Of course, use whatever works best for you! – Mar 15 '16 at 21:05
1

Non-english words with è, for example, will break this. – gavanon Oct 21 '17 at 00:34
@gavanon, yes, that's what I want. – ma11hew28 Oct 21 '17 at 14:30
"abcd".isAlphanumeric giving me // true. But it's not alphanumeric. – Soumen May 23 '18 at 09:54
@Soumen I implemented this property to work like [Python's isalnum](https://docs.python.org/3/library/stdtypes.html#str.isalnum) method, which returns true if the string isn't empty and all of its characters are alphanumeric. – ma11hew28 Jun 18 '18 at 14:10

gavanon · Answer 2 · 2017-10-21T17:02:40.597

30

A modern Swift 3 and 4 solution

extension String {

    func isAlphanumeric() -> Bool {
        return self.rangeOfCharacter(from: CharacterSet.alphanumerics.inverted) == nil && self != ""
    }

    func isAlphanumeric(ignoreDiacritics: Bool = false) -> Bool {
        if ignoreDiacritics {
            return self.range(of: "[^a-zA-Z0-9]", options: .regularExpression) == nil && self != ""
        }
        else {
            return self.isAlphanumeric()
        }
    }

}

Usage:

"".isAlphanumeric()         == false
"Hello".isAlphanumeric()    == true
"Hello 2".isAlphanumeric()  == false
"Hello3".isAlphanumeric()   == true

"Français".isAlphanumeric() == true
"Français".isAlphanumeric(ignoreDiacritics: true) == false

This works with languages other than English, allowing diacritic characters like è and á, etc. If you'd like to ignore these, use the flag "ignoreDiacritics: true".

edited Oct 21 '17 at 17:02

answered Oct 21 '17 at 01:14

gavanon

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This should be the accepted answer. It's cleaner and I suspect it will perform better than any RegEx expression. – R OMS Oct 21 '17 at 14:18
But, this solution allows letters with diacritics, eg, é, which is not what I want. – ma11hew28 Oct 21 '17 at 14:32
@ma11hew28 you should consider allowing for diacritics. Input can unexpectedly have them if user input is anywhere in your code. They are alpha numeric. – gavanon Oct 21 '17 at 14:37
Thank you, @gavanon, I will consider it. But, I am validating usernames for a platform, and I want the users to be able to use the their usernames for their accounts on other platforms, such as Facebook & Instagram, neither of which allow usernames to contain letters with diacritics. – ma11hew28 Oct 21 '17 at 14:49
@ma11hew28 That's to avoid people creating fake accounts like "Mícrosoft" (See what I did there?). I see you updated your question stressing the ignoring of diacritics, so I've included the ability to ignore them in my edited answer. :) – gavanon Oct 21 '17 at 17:08
Nice. Thank you. :-) I think you can delete the first function from your answer and just use its body in the second function. No, sorry, I don't see what you did there. ¯\\_(ツ)_/¯ – ma11hew28 Oct 22 '17 at 04:14
I’ll leave both just for those who want the extra feature. As for Mícrosoft, what I did was use a special í with a diacritic. You didn’t even notice - hence why they’re not allowed in Twitter user names. ;) – gavanon Oct 22 '17 at 14:52
"Hello".isAlphanumeric() == true how this can be true? – Soumen May 23 '18 at 08:49

score 20 · Answer 3 · answered Mar 26 '17 at 16:31

20

I felt the accepted answer using regex was concrete but a rather heavy solution. You could check it this way also in Swift 3.0:

if yourString.rangeOfCharacter(from: CharacterSet.alphanumerics.inverted) != nil {
    return "Username can only contain numbers or digits"
}

answered Mar 26 '17 at 16:31

carbonr

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This feels like a much more sane solution. – Tyten May 19 '17 at 21:07
3

Thank you, but this solution allows letters with diacritics, eg, é, which is not what I want. – ma11hew28 Oct 21 '17 at 14:39
Excellent answer. I think CharacterSet was introduced exactly to get rid of regulars usage – rommex Sep 19 '19 at 11:49

score 7 · Answer 4 · answered Mar 20 '18 at 18:49

extension String {
    /// Allows only `a-zA-Z0-9`
    public var isAlphanumeric: Bool {
        guard !isEmpty else {
            return false
        }
        let allowed = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890"
        let characterSet = CharacterSet(charactersIn: allowed)
        guard rangeOfCharacter(from: characterSet.inverted) == nil else {
            return false
        }
        return true
    }
}

XCTAssertFalse("".isAlphanumeric)
XCTAssertFalse("climate change".isAlphanumeric)
XCTAssertFalse("Poüet".isAlphanumeric)
XCTAssertTrue("Hawking2018".isAlphanumeric)

Checked now and find it's working. I was checking in the playground may be made some mistake then. Thanks. — Soumen, May 23 '18 at 12:12

score 3 · Answer 5 · answered Nov 09 '20 at 20:36

The problem with the CharacterSet.alphanumerics CharacterSet is that it is more permissive than [a-zA-Z0-9]. It contains letters with diacritics, Eastern Arabic numerals, etc.

assert(["e", "E", "3"].allSatisfy({ CharacterSet.alphanumerics.contains($0) }))
assert(["ê", "É", "٣"].allSatisfy({ CharacterSet.alphanumerics.contains($0) }))

You can build your own CharacterSet using only the specific 62 "alphanumeric" characters:

extension CharacterSet {
    
    static var alphanumeric62: CharacterSet {
        return lowercase26.union(uppercase26).union(digits10)
    }
    
    static var lowercase26: CharacterSet { CharacterSet(charactersIn: "a"..."z") }
    static var uppercase26: CharacterSet { CharacterSet(charactersIn: "A"..."Z") }
    static var digits10:    CharacterSet { CharacterSet(charactersIn: "0"..."9") }
    
}

assert(["e", "E", "3"].allSatisfy({ CharacterSet.alphanumeric62.contains($0) }))
assert(["ê", "É", "٣"].allSatisfy({ CharacterSet.alphanumeric62.contains($0) == false }))

Then test your string against the inverse of that CharacterSet:

guard "string".rangeOfCharacter(from: CharacterSet.alphanumeric62.inverted) == nil else {
    fatalError()
}

Amazing, was looking for something like that, thanks! – YYfim May 25 '22 at 06:48 — YYfim, May 25 '22 at 06:48

score 2 · Answer 6 · answered Mar 10 '22 at 15:31

2

Here's a succinct approach:

extension String {
    var isAlphanumeric: Bool {
       allSatisfy { $0.isLetter || $0.isNumber }
    }
}

answered Mar 10 '22 at 15:31

rmp251

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Problem with this approach is it will allow diacritical variants, such as `é`. A workaround is to add `$0.isASCII` to the check. – Ranoiaetep May 12 '22 at 05:33

score 1 · Answer 7 · edited Jan 19 '18 at 05:37

1

This regex used to check that string contains atleast 1 alphabet + atleast 1 digit alongwith 8 characters.

"^(?=.*[A-Za-z])(?=.*\\d)[A-Za-z\\d]{8,}$"

edited Jan 19 '18 at 05:37

Uma Madhavi

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answered Nov 11 '16 at 04:26

Sarishti batta

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score 1 · Answer 8 · answered Jul 17 '20 at 09:57

extension String {  
   var isAlphaNumeric: Bool {
        let hasLetters = rangeOfCharacter(from: .letters, options: .numeric, range: nil) != nil
        let hasNumbers = rangeOfCharacter(from: .decimalDigits, options: .literal, range: nil) != nil
        let comps = components(separatedBy: .alphanumerics)
        return comps.joined(separator: "").count == 0 && hasLetters && hasNumbers  
   } 
}

Check if a String is alphanumeric in Swift

8 Answers8

A modern Swift 3 and 4 solution

Linked