Swift - Encode Url

Swift - encode URL

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)\'")
}

Swift Encode Custom String to populate URL

Your parameters needs to be url encoded.

You can use URLComponents to create your URL and pass your parameters as URLQueryItem instances, like that:

let productSearchString = "iphone 12"
var urlComponents = URLComponents(string: "https://test@api.search")
urlComponents?.queryItems = [
    URLQueryItem(name: "query", value: productSearchString),
    URLQueryItem(name: "limit", value: "1")
]
let url = urlComponents?.url

print(url?.absoluteString ?? "nil") // https://test@api.search?query=iphone%2012&limit=1

URL Encoding of String accepting all special characters in swift

Thank you @MartinR and @benleggiero for : How do I URL encode a string

It helped a lot.

It was not including all special characters.

Checked one by one and added those which were missing as below:

extension CharacterSet {

    public static let urlQueryParameterAllowed = CharacterSet.urlQueryAllowed.subtracting(CharacterSet(charactersIn: "&?~!$*(.,)_-+':"))

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

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


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

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