How do I URL encode a string

Question

I have a URL string (NSString) with spaces and & characters. How do I url encode the entire string (including the & ampersand

Answer 1

Swift 2.0 Example (iOS 9 Compatiable)

extension String {

  func stringByURLEncoding() -> String? {

    let characters = NSCharacterSet.URLQueryAllowedCharacterSet().mutableCopy() as! NSMutableCharacterSet

    characters.removeCharactersInString("&")

    guard let encodedString = self.stringByAddingPercentEncodingWithAllowedCharacters(characters) else {
      return nil
    }

    return encodedString

  }

}

Answer 2

Here's a production-ready flexible approach in Swift 5.x:

public extension CharacterSet {

    static let urlQueryParameterAllowed = CharacterSet.urlQueryAllowed.subtracting(CharacterSet(charactersIn: "&?"))

    static let urlQueryDenied           = CharacterSet.urlQueryAllowed.inverted()
    static let urlQueryKeyValueDenied   = CharacterSet.urlQueryParameterAllowed.inverted()
    static let urlPathDenied            = CharacterSet.urlPathAllowed.inverted()
    static let urlFragmentDenied        = CharacterSet.urlFragmentAllowed.inverted()
    static let urlHostDenied            = CharacterSet.urlHostAllowed.inverted()

    static let urlDenied                = CharacterSet.urlQueryDenied
        .union(.urlQueryKeyValueDenied)
        .union(.urlPathDenied)
        .union(.urlFragmentDenied)
        .union(.urlHostDenied)


    func inverted() -> CharacterSet {
        var copy = self
        copy.invert()
        return copy
    }
}



public extension String {
    func urlEncoded(denying deniedCharacters: CharacterSet = .urlDenied) -> String? {
        return addingPercentEncoding(withAllowedCharacters: deniedCharacters.inverted())
    }
}

Example usage:

print("Hello, World!".urlEncoded()!)
print("You&Me?".urlEncoded()!)
print("#Blessed 100%".urlEncoded()!)
print("Pride and Prejudice".urlEncoded(denying: .uppercaseLetters)!)

Output:

Hello,%20World!
You%26Me%3F
%23Blessed%20100%25
%50ride and %50rejudice

Answer 3

Try to use stringByAddingPercentEncodingWithAllowedCharacters method with [NSCharacterSet URLUserAllowedCharacterSet] it will cover all the cases

Objective C

NSString *value = @"Test / Test";
value = [value stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLUserAllowedCharacterSet]];

swift

var value = "Test / Test"
value.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLUserAllowedCharacterSet())

Output

Test%20%2F%20Test

Answer 4

In my case where the last component was Arabic letters I did the following in Swift 2.2:

extension String {

 func encodeUTF8() -> String? {

    //If I can create an NSURL out of the string nothing is wrong with it
    if let _ = NSURL(string: self) {

        return self
    }

    //Get the last component from the string this will return subSequence
    let optionalLastComponent = self.characters.split { $0 == "/" }.last


    if let lastComponent = optionalLastComponent {

        //Get the string from the sub sequence by mapping the characters to [String] then reduce the array to String
        let lastComponentAsString = lastComponent.map { String($0) }.reduce("", combine: +)


        //Get the range of the last component
        if let rangeOfLastComponent = self.rangeOfString(lastComponentAsString) {
            //Get the string without its last component
            let stringWithoutLastComponent = self.substringToIndex(rangeOfLastComponent.startIndex)


            //Encode the last component
            if let lastComponentEncoded = lastComponentAsString.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.alphanumericCharacterSet()) {


            //Finally append the original string (without its last component) to the encoded part (encoded last component)
            let encodedString = stringWithoutLastComponent + lastComponentEncoded

                //Return the string (original string/encoded string)
                return encodedString
            }
        }
    }

    return nil;
}
}

usage:

let stringURL = "http://xxx.dev.com/endpoint/nonLatinCharacters"

if let encodedStringURL = stringURL.encodeUTF8() {

    if let url = NSURL(string: encodedStringURL) {

      ...
    }

} 

Answer 5

New APIs have been added since the answer was selected; You can now use NSURLUtilities. Since different parts of URLs allow different characters, use the applicable character set. The following example encodes for inclusion in the query string:

encodedString = [myString stringByAddingPercentEncodingWithAllowedCharacters:NSCharacterSet.URLQueryAllowedCharacterSet];

To specifically convert '&', you'll need to remove it from the url query set or use a different set, as '&' is allowed in a URL query:

NSMutableCharacterSet *chars = NSCharacterSet.URLQueryAllowedCharacterSet.mutableCopy;
[chars removeCharactersInRange:NSMakeRange('&', 1)]; // %26
encodedString = [myString stringByAddingPercentEncodingWithAllowedCharacters:chars];

Answer 6

This might be helpful

NSString *sampleUrl = @"http://www.google.com/search.jsp?params=Java Developer";
NSString* encodedUrl = [sampleUrl stringByAddingPercentEscapesUsingEncoding:
 NSUTF8StringEncoding];

For iOS 7+, the recommended way is:

NSString* encodedUrl = [sampleUrl stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLQueryAllowedCharacterSet]];

You can choose the allowed character set as per the requirement of the URL component.