Convert String to NSURL is return nil in swift

Question

I am trying to convert a String to NSURL and my code for that is Below:

var url = \"https://maps.googleapis.com/maps/api/distancema         


        

Answer 1

You can get following error if NSURL is null and trying load http URL over web view:

fatal error: unexpectedly found nil while unwrapping an Optional value

To be safe we should use:

 var urlStr = strLink!.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)

if var url = NSURL(string: urlStr!){
    println(self.strLink!)

    self.webView!.loadRequest(NSURLRequest(URL: url))

}

Answer 2

as blwinters said, in Swift 3.0 use

URL(string: urlPath.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)!)

Answer 3

this work for me

let url : NSString = MyUrls.baseUrl + self.url_file_open as NSString
let urlStr : NSString = url.addingPercentEscapes(using: String.Encoding.utf8.rawValue)! as NSString
if let url = URL(string: urlStr as String) {
    let request = URLRequest(url: url)
    self.businessPlanView.loadRequest(request)
}

Answer 4

I think try this it's perfectly work for me

  var url : String = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=-34.4232722,150.8865837&destinations=-34.4250728,150.89314939999997&language=en-US"
        println("This is String: \(url)")
        var urlStr : NSString = url.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)!
        var remoteUrl : NSURL? = NSURL(string: url)
        println("This is URL: \(remoteUrl!)")

Answer 5

As suggested by the Martin R, I see THIS post and I converted that objective-c code to swift and I got this code:

var url : NSString = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=\(self.latitud‌​e),\(self.longitude)&destinations=\(self.stringForDistance)&language=en-US" 
var urlStr : NSString = url.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)! 
var searchURL : NSURL = NSURL(string: urlStr)! 
println(searchURL)

and this is working correctly.

For swift 3.0:

let url : NSString = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=\(self.latitud‌​e),\(self.longitude)&destinations=\(self.stringForDistance)&language=en-US"
let urlStr : NSString = url.addingPercentEscapes(using: String.Encoding.utf8.rawValue)! as NSString
let searchURL : NSURL = NSURL(string: urlStr as String)!
print(searchURL)

Answer 6

SWIFT 3.0

A safe way to fix a bad string being converted to NSURL is by unwrapping the urlPath string variable using "guard let"

guard let url = NSURL(string: urlPath.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)!)
            else
            {
                print("Couldn't parse myURL = \(urlPath)")
                return
            }

The variable called "urlPath" in my above example would be the url string you have already declared somewhere else in your code.

I came across this answer because randomly I was getting the nil error with XCode breaking at the point my string was made into a NSURL. No logic as to why it was random even when I printed the url's they would look fine. As soon as I added the .addingPercentEncoding it was back working without any issues whatsoever.

tl;dr For anyone reading this, try my above code and swap out "urlPath" for your own local string url.