URL decode in iOS - ios

Question

I am using Swift 1.2 to develop my iPhone application and I am communicating with a http web service.
The response I am getting is in query string format (key-value pairs) and URL encoded in .Net.
I can get the response, but looking the proper way to decode using Swift.
Sample response is as follows
status=1&message=The+transaction+for+GBP+12.50+was+successful
Tried following way to decode and get the server response
// This provides encoded response String
var responseString = NSString(data: data, encoding: NSUTF8StringEncoding) as! String
var decodedResponse = responseString.stringByReplacingEscapesUsingEncoding(NSUTF8StringEncoding)!
How can I replace all URL escaped characters in the string?

Answer 1

To encode and decode urls create this extention somewhere in the project:
Swift 2.0
extension String
{
func encodeUrl() -> String
{
return self.stringByAddingPercentEncodingWithAllowedCharacters( NSCharacterSet.URLQueryAllowedCharacterSet())
}
func decodeUrl() -> String
{
return self.stringByRemovingPercentEncoding
}
}
Swift 3.0
extension String
{
func encodeUrl() -> String
{
return self.addingPercentEncoding( withAllowedCharacters: NSCharacterSet.urlQueryAllowed())
}
func decodeUrl() -> String
{
return self.removingPercentEncoding
}
}
Swift 4.1
extension String
{
func encodeUrl() -> String?
{
return self.addingPercentEncoding( withAllowedCharacters: NSCharacterSet.urlQueryAllowed)
}
func decodeUrl() -> String?
{
return self.removingPercentEncoding
}
}

Answer 2

Swift 2 and later (Xcode 7)
var s = "aa bb -[:/?&=;+!##$()',*]";
let sEncode = s.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLQueryAllowedCharacterSet())
let sDecode = sEncode?.stringByRemovingPercentEncoding

Answer 3

You only need:
print("Decode: ", yourUrlAsString.removingPercentEncoding)

Answer 4

The stringByReplacingEscapesUsingEncoding method is behaving correctly. The "+" character is not part of percent-encoding. This server is using it incorrectly; it should be using a percent-escaped space here (%20). If, for a particular response, you want spaces where you see "+" characters, you just have to work around the server behavior by performing the substitution yourself, as you are already doing.

Answer 5

It's better to use built-in URLComponents struct, since it follows proper guidelines.
extension URL
{
var parameters: [String: String?]?
{
if let components = URLComponents(url: self, resolvingAgainstBaseURL: false),
let queryItems = components.queryItems
{
var parameters = [String: String?]()
for item in queryItems {
parameters[item.name] = item.value
}
return parameters
} else {
return nil
}
}
}

Answer 6

In my case, I NEED a plus ("+") signal in a phone number in parameters of a query string, like "[phone]". After I discovered that the swift3 (xcode 8.2) encoder don't encode "+" as plus signal, but space, I had to appeal to a workaround after the encode:
Swift 3.0
_strURL = _strURL.replacingOccurrences(of: "+", with: "%2B")

Answer 7

In Swift 3
extension URL {
var parseQueryString: [String: String] {
var results = [String: String]()
if let pairs = self.query?.components(separatedBy: "&"), pairs.count > 0 {
for pair: String in pairs {
if let keyValue = pair.components(separatedBy: "=") as [String]? {
results.updateValue(keyValue[1], forKey: keyValue[0])
}
}
}
return results
}
}
in your code to access below
let parse = url.parseQueryString
print("parse \(parse)" )