Check if string contains special characters in Swift

Question

I have to detect whether a string contains any special characters. How can I check it? Does Swift support regular expressions?

var characterSet:NSCharacterSet = NSCharacterSet(charactersInString: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLKMNOPQRSTUVWXYZ0123456789")
if (searchTerm!.rangeOfCharacterFromSet(characterSet).location == NSNotFound){
    println("Could not handle special characters")
}

I tried the code above, but it matches only if I enter the first character as a special character.

Answer 1

Your code check if no character in the string is from the given set. What you want is to check if any character is not in the given set:

if (searchTerm!.rangeOfCharacterFromSet(characterSet.invertedSet).location != NSNotFound){
    println("Could not handle special characters")
}

You can also achieve this using regular expressions:

let regex = NSRegularExpression(pattern: ".*[^A-Za-z0-9].*", options: nil, error: nil)!
if regex.firstMatchInString(searchTerm!, options: nil, range: NSMakeRange(0, searchTerm!.length)) != nil {
    println("could not handle special characters")

}

The pattern [^A-Za-z0-9] matches a character which is not from the ranges A-Z, a-z, or 0-9.

Update for Swift 2:

let searchTerm = "a+b"

let characterset = NSCharacterSet(charactersInString: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789")
if searchTerm.rangeOfCharacterFromSet(characterset.invertedSet) != nil {
    print("string contains special characters")
}

Update for Swift 3:

let characterset = CharacterSet(charactersIn: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789")
if searchTerm.rangeOfCharacter(from: characterset.inverted) != nil {
    print("string contains special characters")
}

Answer 2

This answer may help the people who are using Swift 4.1

func hasSpecialCharacters() -> Bool {

    do {
        let regex = try NSRegularExpression(pattern: ".*[^A-Za-z0-9].*", options: .caseInsensitive)
        if let _ = regex.firstMatch(in: self, options: NSRegularExpression.MatchingOptions.reportCompletion, range: NSMakeRange(0, self.count)) {
            return true
        }

    } catch {
        debugPrint(error.localizedDescription)
        return false
    }

    return false
}

Taken reference from @Martin R's answer.

Answer 3

Inverting your character set will work, because in your character set you have all the valid characters:

var characterSet:NSCharacterSet = NSCharacterSet(charactersInString: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLKMNOPQRSTUVWXYZ0123456789")
if (searchTerm!.rangeOfCharacterFromSet(characterSet.invertedSet).location == NSNotFound){
    println("No special characters")
}

Hope this helps.. :)

Answer 4

@Martin R answer is great, I just wanted to update it (the second part) to Swift 2.1 version

let regex = try! NSRegularExpression(pattern: ".*[^A-Za-z0-9].*", options: NSRegularExpressionOptions())
if regex.firstMatchInString(searchTerm!, options: NSMatchingOptions(), range:NSMakeRange(0, searchTerm!.characters.count)) != nil {
    print("could not handle special characters")
}

I used try! as we can be sure it create a regex, it doesn't base on any dynamic kind of a data

Answer 5

Password validation With following:- (Password at least eight characters long, one special character, one uppercase, one lower case letter and one digit)

var isValidateSecialPassword : Bool {

        if(self.count>=8 && self.count<=20){
        }else{
            return false
        }
        let nonUpperCase = CharacterSet(charactersIn: "ABCDEFGHIJKLMNOPQRSTUVWXYZ").inverted
        let letters = self.components(separatedBy: nonUpperCase)
        let strUpper: String = letters.joined()

        let smallLetterRegEx  = ".*[a-z]+.*"
        let samlltest = NSPredicate(format:"SELF MATCHES %@", smallLetterRegEx)
        let smallresult = samlltest.evaluate(with: self)

        let numberRegEx  = ".*[0-9]+.*"
        let numbertest = NSPredicate(format:"SELF MATCHES %@", numberRegEx)
        let numberresult = numbertest.evaluate(with: self)

        let regex = try! NSRegularExpression(pattern: ".*[^A-Za-z0-9].*", options: NSRegularExpression.Options())
        var isSpecial :Bool = false
        if regex.firstMatch(in: self, options: NSRegularExpression.MatchingOptions(), range:NSMakeRange(0, self.count)) != nil {
            print("could not handle special characters")
            isSpecial = true
        }else{
            isSpecial = false
        }
        return (strUpper.count >= 1) && smallresult && numberresult && isSpecial
    }

Answer 6

Depending on the definition of special characters, you could use this:

let chars =  "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLKMNOPQRSTUVWXYZ0123456789"

chars.canBeConvertedToEncoding(NSASCIIStringEncoding)

Answer 7

Two Solutions:

1)

extension String {
    var stripped: String {
        let okayChars = Set("abcdefghijklmnopqrstuvwxyz ABCDEFGHIJKLKMNOPQRSTUVWXYZ")
        return self.filter {okayChars.contains($0) }
    }
}

2)

class TrimDictionary {

    static func trimmedWord(wordString: String) -> String {
        var selectedString = wordString

        let strFirst = selectedString.first
        let strLast = selectedString.last

        let characterset = CharacterSet(charactersIn: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
        if strFirst?.description.rangeOfCharacter(from: characterset.inverted) != nil {
            selectedString = String(selectedString.dropFirst())
        }
        if strLast?.description.rangeOfCharacter(from: characterset.inverted) != nil {
            selectedString = String(selectedString.dropLast())
        }
        return selectedString
    }
}