Conversion stringByAddingPercentEncodingWithAllowedCharacters(.URLHostAllowedCharacterSet()) from Swift2 to Swift3 [duplicate] - swift

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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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;
}
}

Answer 6

Swift 4.0
let encodedData = myUrlString.addingPercentEncoding(withAllowedCharacters: CharacterSet.urlHostAllowed)