URL decode in 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?

Best to update to Swift 2 now rather than later. And it really is better and easier to code in. — zaph, Oct 06 '15 at 16:22
"replace `+` with `" "`, before use `stringByReplacingEscapesUsingEncoding` method" So what's the problem? Do that and see what happens. — matt, Oct 06 '15 at 16:22
@matt, Yes, I've done it and + gets replaced with Space. But wondering if any other special cases (Like +) can be there and will NOT read the exact thing the server sends — JibW, Oct 06 '15 at 16:25

score 104 · Answer 1 · edited Apr 12 '18 at 07:34

104

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
    }
}

edited Apr 12 '18 at 07:34

Kushal Ashok

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answered Sep 14 '16 at 12:29

Vyacheslav

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`addingPercentEncoding` and `removingPercentEncoding` return `Optional` strings so the return type should be `String?`. – Extragorey May 26 '17 at 01:41
Even simpler than the [Apple's guide](https://developer.apple.com/library/content/documentation/Cocoa/Conceptual/URLLoadingSystem/WorkingwithURLEncoding/WorkingwithURLEncoding.html). Thanks. – Vadim Bulavin Feb 01 '18 at 11:38
3

This is very incorrect. The character set constants were not made to encode URL, and for that reason you cannot use any of the predefined sets to encode URL. You would have to create a new one. Your code won't encode `&` and `=` correctly, for example. Decoding will work though. – Sulthan Mar 13 '18 at 16:32
decoding string by removingPercentEncoding after that string converted %5Cn to \n but not working \n in UILabel and Hard code string \n is working in UIlabel showing by new line but decoded string not working. can you please help me to give suggestion for this? – nipa Dec 11 '20 at 03:28

score 9 · Answer 2 · edited Apr 27 '20 at 19:41

9

Swift 2 and later (Xcode 7)

var s = "aa bb -[:/?&=;+!@#$()',*]";

let sEncode = s.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLQueryAllowedCharacterSet())

let sDecode = sEncode?.stringByRemovingPercentEncoding

edited Apr 27 '20 at 19:41

shim

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answered Oct 22 '15 at 06:01

米米米

970
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score 9 · Answer 3 · answered May 29 '20 at 21:48

9

You only need:

print("Decode: ", yourUrlAsString.removingPercentEncoding)

answered May 29 '20 at 21:48

ScottyBlades

12,189
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matt · Answer 4 · 2015-10-06T16:42:09.063

7

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.

edited Oct 06 '15 at 16:42

answered Oct 06 '15 at 16:37

matt

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And see this answer to understand what the `+` is really for: http://stackoverflow.com/a/2700981/341994 – matt Oct 06 '15 at 16:41
2

Hi matt, yes I know the server use + instead of %20 to represent space in the response (encoded query string response)... Java handles this situation and decode properly "URLDecoder.decode(responseString, "UTF-8");"... I have added my temporary way made it working under edits in my question, but looking if there is any straight forward language API method available to decode properly rather than me replacing + with space before perform stringByReplacingEscapesUsingEncoding... Are you saying if that type of method NOT available in SWIFT? – JibW Oct 07 '15 at 08:49

score 3 · Answer 5 · answered Oct 19 '19 at 11:30

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
        }
    }
}

score 0 · Answer 6 · answered Dec 21 '16 at 22:54

In my case, I NEED a plus ("+") signal in a phone number in parameters of a query string, like "+55 11 99999-5555". 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")

score 0 · Answer 7 · answered Jan 05 '18 at 05:29

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)" )

URL decode in iOS

7 Answers7

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