Swift - encode URL

IosSwiftUrlencode

Ios Problem Overview

If I encode a string like this:

var escapedString = originalString.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)

it doesn't escape the slashes /.

I've searched and found this Objective C code:

NSString *encodedString = (NSString *)CFURLCreateStringByAddingPercentEscapes(
                        NULL,
                        (CFStringRef)unencodedString,
                        NULL,
                        (CFStringRef)@"!*'();:@&=+$,/?%#[]",
                        kCFStringEncodingUTF8 );

Is there an easier way to encode an URL and if not, how do I write this in Swift?

Ios Solutions

Solution 1 - Ios

Swift 3

In Swift 3 there is addingPercentEncoding

let originalString = "test/test"
let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
print(escapedString!)

Output:

> test%2Ftest

Swift 1

In iOS 7 and above there is stringByAddingPercentEncodingWithAllowedCharacters

var originalString = "test/test"
var escapedString = originalString.stringByAddingPercentEncodingWithAllowedCharacters(.URLHostAllowedCharacterSet())
println("escapedString: \(escapedString)")

Output:

> test%2Ftest

The following are useful (inverted) character sets:

URLFragmentAllowedCharacterSet  "#%<>[\]^`{|}
URLHostAllowedCharacterSet      "#%/<>?@\^`{|}
URLPasswordAllowedCharacterSet  "#%/:<>?@[\]^`{|}
URLPathAllowedCharacterSet      "#%;<>?[\]^`{|}
URLQueryAllowedCharacterSet     "#%<>[\]^`{|}
URLUserAllowedCharacterSet      "#%/:<>?@[\]^`

If you want a different set of characters to be escaped create a set:
Example with added "=" character:

var originalString = "test/test=42"
var customAllowedSet =  NSCharacterSet(charactersInString:"=\"#%/<>?@\\^`{|}").invertedSet
var escapedString = originalString.stringByAddingPercentEncodingWithAllowedCharacters(customAllowedSet)
println("escapedString: \(escapedString)")

Output:

> test%2Ftest%3D42

Example to verify ascii characters not in the set:

func printCharactersInSet(set: NSCharacterSet) {
    var characters = ""
    let iSet = set.invertedSet
    for i: UInt32 in 32..<127 {
        let c = Character(UnicodeScalar(i))
        if iSet.longCharacterIsMember(i) {
            characters = characters + String(c)
        }
    }
    print("characters not in set: \'\(characters)\'")
}

Solution 2 - Ios

You can use URLComponents to avoid having to manually percent encode your query string:

let scheme = "https"
let host = "www.google.com"
let path = "/search"
let queryItem = URLQueryItem(name: "q", value: "Formula One")


var urlComponents = URLComponents()
urlComponents.scheme = scheme
urlComponents.host = host
urlComponents.path = path
urlComponents.queryItems = [queryItem]

if let url = urlComponents.url {
    print(url)   // "https://www.google.com/search?q=Formula%20One"
}

extension URLComponents {
    init(scheme: String = "https",
         host: String = "www.google.com",
         path: String = "/search",
         queryItems: [URLQueryItem]) {
        self.init()
        self.scheme = scheme
        self.host = host
        self.path = path
        self.queryItems = queryItems
    }
}

let query = "Formula One"
if let url = URLComponents(queryItems: [URLQueryItem(name: "q", value: query)]).url {
    print(url)  // https://www.google.com/search?q=Formula%20One
}

Solution 3 - Ios

Swift 4 & 5

To encode a parameter in URL I find using .alphanumerics character set the easiest option:

let urlEncoded = value.addingPercentEncoding(withAllowedCharacters: .alphanumerics)
let url = "http://www.example.com/?name=\(urlEncoded!)"

Using any of the standard Character Sets for URL Encoding (like URLQueryAllowedCharacterSet or URLHostAllowedCharacterSet) won't work, because they do not exclude = or & characters.

Note that by using .alphanumerics it will encode some characters that do not need to be encoded (like -, ., _ or ~ -– see 2.3. Unreserved characters in RFC 3986). I find using .alphanumerics simpler than constructing a custom character set and do not mind some additional characters to be encoded. If that bothers you, construct a custom character set as is described in How to percent encode a URL String, like for example:

// Store allowed character set for reuse (computed lazily).
private let urlAllowed: CharacterSet =
    .alphanumerics.union(.init(charactersIn: "-._~")) // as per RFC 3986

extension String {
    var urlEncoded: String? {
        return addingPercentEncoding(withAllowedCharacters: urlAllowed)
    }
}

let url = "http://www.example.com/?name=\(value.urlEncoded!)"

Warning: The urlEncoded parameter is force unwrapped. For invalid unicode string it might crash. See https://stackoverflow.com/questions/33558933/why-is-the-return-value-of-string-addingpercentencoding-optional. Instead of force unwrapping urlEncoded! you can use urlEncoded ?? "" or if let urlEncoded = urlEncoded { ... }.

Solution 4 - Ios

Swift 5:

extension String {
    var urlEncoded: String? {
        let allowedCharacterSet = CharacterSet.alphanumerics.union(CharacterSet(charactersIn: "~-_."))
        return self.addingPercentEncoding(withAllowedCharacters: allowedCharacterSet)
    }
}

print("\u{48}ello\u{9}world\u{7}\u{0}".urlEncoded!) // prints Hello%09world%07%00
print("The string ü@foo-bar".urlEncoded!) // prints The%20string%20%C3%BC%40foo-bar

Solution 5 - Ios

Swift 3:
let originalString = "http://www.ihtc.cc?name=htc&title=iOS开发工程师"

1. encodingQuery:

let escapedString = originalString.addingPercentEncoding(withAllowedCharacters:NSCharacterSet.urlQueryAllowed)

result:

"http://www.ihtc.cc?name=htc&title=iOS%E5%BC%80%E5%8F%91%E5%B7%A5%E7%A8%8B%E5%B8%88"

2. encodingURL:

let escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)

result:

"http:%2F%2Fwww.ihtc.cc%3Fname=htc&title=iOS%E5%BC%80%E5%8F%91%E5%B7%A5%E7%A8%8B%E5%B8%88"

Solution 6 - Ios

Swift 4 & 5 (Thanks @sumizome for suggestion. Thanks @FD_ and @derickito for testing)

var allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed
allowedQueryParamAndKey.remove(charactersIn: ";/?:@&=+$, ")
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)

Swift 3

let allowedQueryParamAndKey =  NSCharacterSet.urlQueryAllowed.remove(charactersIn: ";/?:@&=+$, ")
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)

Swift 2.2 (Borrowing from Zaph's and correcting for url query key and parameter values)

var allowedQueryParamAndKey =  NSCharacterSet(charactersInString: ";/?:@&=+$, ").invertedSet
paramOrKey.stringByAddingPercentEncodingWithAllowedCharacters(allowedQueryParamAndKey)

Example:

let paramOrKey = "https://some.website.com/path/to/page.srf?a=1&b=2#top"
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
// produces:
"https%3A%2F%2Fsome.website.com%2Fpath%2Fto%2Fpage.srf%3Fa%3D1%26b%3D2%23top"

This is a shorter version of Bryan Chen's answer. I'd guess that urlQueryAllowed is allowing the control characters through which is fine unless they form part of the key or value in your query string at which point they need to be escaped.

Solution 7 - Ios

Swift 4:

It depends by the encoding rules followed by your server.

Apple offer this class method, but it don't report wich kind of RCF protocol it follows.

var escapedString = originalString.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)!

Following this useful tool you should guarantee the encoding of these chars for your parameters:

  • $ (Dollar Sign) becomes %24
  • & (Ampersand) becomes %26
  • + (Plus) becomes %2B
  • , (Comma) becomes %2C
  • : (Colon) becomes %3A
  • ; (Semi-Colon) becomes %3B
  • = (Equals) becomes %3D
  • ? (Question Mark) becomes %3F
  • @ (Commercial A / At) becomes %40

In other words, speaking about URL encoding, you should following the RFC 1738 protocol.

And Swift don't cover the encoding of the + char for example, but it works well with these three @ : ? chars.

So, to correctly encoding each your parameter , the .urlHostAllowed option is not enough, you should add also the special chars as for example:

encodedParameter = parameter.replacingOccurrences(of: "+", with: "%2B")

Hope this helps someone who become crazy to search these informations.

Solution 8 - Ios

Everything is same

var str = CFURLCreateStringByAddingPercentEscapes(
    nil,
    "test/test",
    nil,
    "!*'();:@&=+$,/?%#[]",
    CFStringBuiltInEncodings.UTF8.rawValue
)

// test%2Ftest

Solution 9 - Ios

Swift 4.2

A quick one line solution. Replace originalString with the String you want to encode.

var encodedString = originalString.addingPercentEncoding(withAllowedCharacters: CharacterSet(charactersIn: "!*'();:@&=+$,/?%#[]{} ").inverted)

Online Playground Demo

Solution 10 - Ios

Had need of this myself, so I wrote a String extension that both allows for URLEncoding strings, as well as the more common end goal, converting a parameter dictionary into "GET" style URL Parameters:

extension String {
    func URLEncodedString() -> String? {
        var escapedString = self.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
        return escapedString
    }
    static func queryStringFromParameters(parameters: Dictionary<String,String>) -> String? {
        if (parameters.count == 0)
        {
            return nil
        }
        var queryString : String? = nil
        for (key, value) in parameters {
            if let encodedKey = key.URLEncodedString() {
                if let encodedValue = value.URLEncodedString() {
                    if queryString == nil
                    {
                        queryString = "?"
                    }
                    else
                    {
                        queryString! += "&"
                    }
                    queryString! += encodedKey + "=" + encodedValue
                }
            }
        }
        return queryString
    }
}

Enjoy!

Solution 11 - Ios

This is working for me in Swift 5. The usage case is taking a URL from the clipboard or similar which may already have escaped characters but which also contains Unicode characters which could cause URLComponents or URL(string:) to fail.

First, create a character set that includes all URL-legal characters:

extension CharacterSet {
    
    /// Characters valid in at least one part of a URL.
    ///
    /// These characters are not allowed in ALL parts of a URL; each part has different requirements. This set is useful for checking for Unicode characters that need to be percent encoded before performing a validity check on individual URL components.
    static var urlAllowedCharacters: CharacterSet {
        // Start by including hash, which isn't in any set
        var characters = CharacterSet(charactersIn: "#")
        // All URL-legal characters
        characters.formUnion(.urlUserAllowed)
        characters.formUnion(.urlPasswordAllowed)
        characters.formUnion(.urlHostAllowed)
        characters.formUnion(.urlPathAllowed)
        characters.formUnion(.urlQueryAllowed)
        characters.formUnion(.urlFragmentAllowed)
        
        return characters
    }
}

Next, extend String with a method to encode URLs:

extension String {
    
    /// Converts a string to a percent-encoded URL, including Unicode characters.
    ///
    /// - Returns: An encoded URL if all steps succeed, otherwise nil.
    func encodedUrl() -> URL? {        
        // Remove preexisting encoding,
        guard let decodedString = self.removingPercentEncoding,
            // encode any Unicode characters so URLComponents doesn't choke,
            let unicodeEncodedString = decodedString.addingPercentEncoding(withAllowedCharacters: .urlAllowedCharacters),
            // break into components to use proper encoding for each part,
            let components = URLComponents(string: unicodeEncodedString),
            // and reencode, to revert decoding while encoding missed characters.
            let percentEncodedUrl = components.url else {
            // Encoding failed
            return nil
        }
        
        return percentEncodedUrl
    }

}

Which can be tested like:

let urlText = "https://www.example.com/폴더/search?q=123&foo=bar&multi=eggs+and+ham&hangul=한글&spaced=lovely%20spam&illegal=<>#top"
let url = encodedUrl(from: urlText)

Value of url at the end: https://www.example.com/%ED%8F%B4%EB%8D%94/search?q=123&foo=bar&multi=eggs+and+ham&hangul=%ED%95%9C%EA%B8%80&spaced=lovely%20spam&illegal=%3C%3E#top

Note that both %20 and + spacing are preserved, Unicode characters are encoded, the %20 in the original urlText is not double encoded, and the anchor (fragment, or #) remains.

Edit: Now checking for validity of each component.

Solution 12 - Ios

For Swift 5 to endcode string

func escape(string: String) -> String {
    let allowedCharacters = string.addingPercentEncoding(withAllowedCharacters: CharacterSet(charactersIn: ":=\"#%/<>?@\\^`{|}").inverted) ?? ""
    return allowedCharacters
}

How to use ?

let strEncoded = self.escape(string: "http://www.edamam.com/ontologies/edamam.owl#recipe_e2a1b9bf2d996cbd9875b80612ed9aa4")
print("escapedString: \(strEncoded)")

Solution 13 - Ios

This one is working for me.

func stringByAddingPercentEncodingForFormData(plusForSpace: Bool=false) -> String? {

    let unreserved = "*-._"
    let allowed = NSMutableCharacterSet.alphanumericCharacterSet()
    allowed.addCharactersInString(unreserved)
    
    if plusForSpace {
        allowed.addCharactersInString(" ")
    }
    
    var encoded = stringByAddingPercentEncodingWithAllowedCharacters(allowed)

    if plusForSpace {
        encoded = encoded?.stringByReplacingOccurrencesOfString(" ", withString: "+")
    }
    return encoded
}

I found above function from this link: http://useyourloaf.com/blog/how-to-percent-encode-a-url-string/.

Solution 14 - Ios

let Url = URL(string: urlString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) ?? "")

Solution 15 - Ios

What helped me was that I created a separate NSCharacterSet and used it on an UTF-8 encoded string i.e. textToEncode to generate the required result:

var queryCharSet = NSCharacterSet.urlQueryAllowed
queryCharSet.remove(charactersIn: "+&?,:;@+=$*()")
    
let utfedCharacterSet = String(utf8String: textToEncode.cString(using: .utf8)!)!
let encodedStr = utfedCharacterSet.addingPercentEncoding(withAllowedCharacters: queryCharSet)!
    
let paramUrl = "https://api.abc.eu/api/search?device=true&query=\(escapedStr)"

Solution 16 - Ios

None of these answers worked for me. Our app was crashing when a url contained non-English characters.

 let unreserved = "-._~/?%$!:"
 let allowed = NSMutableCharacterSet.alphanumeric()
     allowed.addCharacters(in: unreserved)

 let escapedString = urlString.addingPercentEncoding(withAllowedCharacters: allowed as CharacterSet)

Depending on the parameters of what you are trying to do, you may want to just create your own character set. The above allows for english characters, and -._~/?%$!:

Solution 17 - Ios

SWIFT 4.2

Sometimes this happened just because there is space in slug OR absence of URL encoding for parameters passing through API URL.

let myString = self.slugValue
                let csCopy = CharacterSet(bitmapRepresentation: CharacterSet.urlPathAllowed.bitmapRepresentation)
                let escapedString = myString!.addingPercentEncoding(withAllowedCharacters: csCopy)!
                //always "info:hello%20world"
                print(escapedString)

NOTE : Don't forget to explore about bitmapRepresentation.

Solution 18 - Ios

version:Swift 5

// space convert to +
let mstring = string.replacingOccurrences(of: " ", with: "+")
// remove special character
var allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed
allowedQueryParamAndKey.remove(charactersIn: "!*'\"();:@&=+$,/?%#[]%")
return mstring.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey) ?? mstring

Solution 19 - Ios

Swift 5 You can try .afURLQueryAllowed option if you want to encode string like below

let testString = "6hAD9/RjY+SnGm&B"
let escodedString = testString.addingPercentEncoding(withAllowedCharacters: .afURLQueryAllowed)
print(escodedString!)

//encoded string will be like en6hAD9%2FRjY%2BSnGm%26B

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