Check if a String is alphanumeric in Swift

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".

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()
}

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  
   } 
}

Here's a succinct approach:

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