URL（string :)给出nil错误，即使在Swift中String不是nil

Question

I have the following code:

print("CODE STRING SELECTED: \(codeString)")
let aURL = URL(string: codeString)!
if UIApplication.shared.canOpenURL(aURL) { UIApplication.shared.openURL(aURL) }

This code is inside a Button and the Xcode console prints the codeString correctly, it's not nil and so it should open the URL of the codeString , instead, Xcode throws this error:

CODE STRING SELECTED: mailto:john@doe.com?subject=Subject here&body=Lorem ipsum dolor sit, amet quatum.

Fatal error: Unexpectedly found nil while unwrapping an Optional value
2019-07-08 11:19:56.076467+0200 QRcode[2751:1043394] Fatal error: Unexpectedly found nil while unwrapping an Optional value

The same thing happens in case of a Phone number or SMS string (I get the codeString value from a scanned QR code):

CODE STRING SELECTED: tel:+1 2345678901
Fatal error: Unexpectedly found nil while unwrapping an Optional value

CODE STRING SELECTED: SMSTO:+1012345678:lorem ipsum dolor sit, amet
Fatal error: Unexpectedly found nil while unwrapping an Optional value

In case of a URL, like https//example.com, the app doesn't crash, no nil error, same for text, etc. So I really don't understand why I get that error even if the codeString is not nil .

Answer 1

The string is not nil but it does not represent a valid URL. You have to encode the URL.

However it's recommended to unwrap the optionals safely

if let encodedString = codeString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed),
   let aURL = URL(string: encodedString), UIApplication.shared.canOpenURL(aURL) { 
    UIApplication.shared.openURL(aURL) 
}

Answer 2

The url is nil because it cannot be created without escaping spaces you got there.

This would work:

guard let escapedString = codeString.addingPercentEncoding(withAllowedCharacters: urlQuoeryAllowed), 
      let url = URL(string: escapedString) else {
 return
}