Convert String to NSURL is return nil in swift

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 NSStringlet searchURL : NSURL = NSURL(string: urlStr as String)!print(searchURL)

as blwinters said, in Swift 3.0 use

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

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!)")