How to use stringByAddingPercentEncodingWithAllowedCharacters() for a URL in Swift 2.0

Question

I was using this, in Swift 1.2

let urlwithPercentEscapes = myurlstring.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)

This now gives me a warning asking me to use

stringByAddingPercentEncodingWithAllowedCharacters

I need to use a NSCharacterSet as an argument, but there are so many and I cannot determine what one will give me the same outcome as the previously used method.

An example URL I want to use will be like this

http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red Lion&location=19036&location=1090 N Charlotte St, Lancaster, PA

The URL Character Set for encoding seems to contain sets the trim my URL. i.e,

The path component of a URL is the component immediately following the host component (if present). It ends wherever the query or fragment component begins. For example, in the URL http://www.example.com/index.php?key1=value1, the path component is /index.php.

However I don't want to trim any aspect of it. When I used my String, for example myurlstring it would fail.

But when used the following, then there were no issues. It encoded the string with some magic and I could get my URL data.

let urlwithPercentEscapes = myurlstring.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)

As it

Returns a representation of the String using a given encoding to determine the percent escapes necessary to convert the String into a legal URL string

Thanks

score 159 · Accepted Answer · edited Aug 25 '16 at 03:38

For the given URL string the equivalent to

let urlwithPercentEscapes = myurlstring.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)

is the character set URLQueryAllowedCharacterSet

let urlwithPercentEscapes = myurlstring.stringByAddingPercentEncodingWithAllowedCharacters( NSCharacterSet.URLQueryAllowedCharacterSet())

Swift 3:

let urlwithPercentEscapes = myurlstring.addingPercentEncoding( withAllowedCharacters: .urlQueryAllowed)

It encodes everything after the question mark in the URL string.

Since the method stringByAddingPercentEncodingWithAllowedCharacters can return nil, use optional bindings as suggested in the answer of Leo Dabus.

Makes sense, thanks - if you get time can you help with this http://stackoverflow.com/questions/32061298/could-geocodeaddressdictionary-in-ios-9-be-implemented-different-from-ios-8 — DogCoffee, Aug 18 '15 at 07:26

Leo Dabus · Answer 2 · 2018-07-09T10:48:30.583

It will depend on your url. If your url is a path you can use the character set urlPathAllowed

let myFileString = "My File.txt"
if let urlwithPercentEscapes = myFileString.addingPercentEncoding(withAllowedCharacters: .urlPathAllowed) {
    print(urlwithPercentEscapes)  // "My%20File.txt"
}

Creating a Character Set for URL Encoding

urlFragmentAllowed

urlHostAllowed

urlPasswordAllowed

urlQueryAllowed

urlUserAllowed

You can create also your own url character set:

let myUrlString = "http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red Lion&location=19036&location=1090 N Charlotte St, Lancaster, PA"

let urlSet = CharacterSet.urlFragmentAllowed
                .union(.urlHostAllowed)
                .union(.urlPasswordAllowed)
                .union(.urlQueryAllowed)
                .union(.urlUserAllowed)

extension CharacterSet {
    static let urlAllowed = CharacterSet.urlFragmentAllowed
                                        .union(.urlHostAllowed)
                                        .union(.urlPasswordAllowed)
                                        .union(.urlQueryAllowed)
                                        .union(.urlUserAllowed)
}

if let urlwithPercentEscapes = myUrlString.addingPercentEncoding(withAllowedCharacters: .urlAllowed) {
    print(urlwithPercentEscapes)  // "http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red%20Lion&location=19036&location=1090%20N%20Charlotte%20St,%20Lancaster,%20PA"
}

Another option is to use URLComponents to properly create your url

More importantly you need to apply the appropriate character set to each part of the URL, there is no single solution as the different parts have different capabilities... — Wain, Aug 18 '15 at 06:43
Doesn't work, for example the set you suggested returns this, http%3A//www.mapquestapi.com/geocoding/v1/batch%3Fkey= but the older method that works, gave me http://www.mapquestapi.com/geocoding/v1/batch?key= — DogCoffee, Aug 18 '15 at 06:48
@Wain very true, thats what the warning says as well. Using the set in this answer in only the parts the have WhiteSpace seems to work. — DogCoffee, Aug 18 '15 at 07:16
@LeoDabus wouldn't know where to start with that. But will consider it if things go south. On another note, map quest returns some weird syntax around their json. — DogCoffee, Aug 18 '15 at 08:16

score 7 · Answer 3 · edited Apr 02 '18 at 09:50

Swift 3.0 (From grokswift)

Creating URLs from strings is a minefield for bugs. Just miss a single / or accidentally URL encode the ? in a query and your API call will fail and your app won’t have any data to display (or even crash if you didn’t anticipate that possibility). Since iOS 8 there’s a better way to build URLs using NSURLComponents and NSURLQueryItems.

func createURLWithComponents() -> URL? {
        var urlComponents = URLComponents()
        urlComponents.scheme = "http"
        urlComponents.host = "www.mapquestapi.com"
        urlComponents.path = "/geocoding/v1/batch"

        let key = URLQueryItem(name: "key", value: "YOUR_KEY_HERE")
        let callback = URLQueryItem(name: "callback", value: "renderBatch")
        let locationA = URLQueryItem(name: "location", value: "Pottsville,PA")
        let locationB = URLQueryItem(name: "location", value: "Red Lion")
        let locationC = URLQueryItem(name: "location", value: "19036")
        let locationD = URLQueryItem(name: "location", value: "1090 N Charlotte St, Lancaster, PA")

        urlComponents.queryItems = [key, callback, locationA, locationB, locationC, locationD]

        return urlComponents.url
}

Below is the code to access url using guard statement.

guard let url = createURLWithComponents() else {
            print("invalid URL")
            return nil
      }
      print(url)

Output:

http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red%20Lion&location=19036&location=1090%20N%20Charlotte%20St,%20Lancaster,%20PA

score 3 · Answer 4 · answered Jun 08 '17 at 22:58

In Swift 3.1, I am using something like the following:

let query = "param1=value1&param2=" + valueToEncode.addingPercentEncoding(withAllowedCharacters: .alphanumeric)

It's safer than .urlQueryAllowed and the others, because it this will encode every characters other than A-Z, a-z and 0-9. This works better when the value you are encoding may use special characters like ?, &, =, + and spaces.

score 2 · Answer 5 · answered Mar 21 '16 at 19:40

In my case where the last component was non latin characters 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;
}
}

score 0 · Answer 6 · answered Apr 02 '18 at 09:46

0

Swift 4.0

let encodedData = myUrlString.addingPercentEncoding(withAllowedCharacters: CharacterSet.urlHostAllowed)

answered Apr 02 '18 at 09:46

Supertecnoboff

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How to use stringByAddingPercentEncodingWithAllowedCharacters() for a URL in Swift 2.0

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