CODEWITHSUNDEEP
iosswifturl
Chris
Chris

Reputation: 2274

Correct way to handle incorrect URL when opening

The task is quite simple. With a given urlString open it when it is valid. This is what I tried:

func openURL(_ urlString: String) {
    guard let url = URL(string: urlString) else {
        showInvalidUrlAlert()
        return
    }
    UIApplication.shared.open(url)
}

This work with this example: "https://www.google.de/?hl=de"

However when passing an invalid url, which is also possible in my application (for example: "asdfd") I get this error on the console but nothing happens in the app:

[default] Failed to open URL asdf: Error Domain=NSOSStatusErrorDomain Code=-50 "invalid input parameters" UserInfo={NSDebugDescription=invalid input parameters, _LSLine=252, _LSFunction=-[_LSDOpenClient openURL:options:completionHandler:]}

What is the best practice here?

Upvotes: 2

Views: 7805

Answers (3)

CDM
CDM

Reputation: 517

Did the URL have a http or https scheme attached? to open a website, you must require the String to have http(s). other wise the application won't know how to handle it, since it also handles other protocols.

...supports many common schemes, including the http, https, tel, facetime, and mailto schemes...

https://developer.apple.com/documentation/uikit/uiapplication/1648685-open

Upvotes: 6

pawello2222
pawello2222

Reputation: 54466

You may want to use the completionHandler parameter:

func openURL(_ urlString: String) {
    guard let url = URL(string: urlString) else {
        showInvalidUrlAlert()
        return
    }
    UIApplication.shared.open(url, completionHandler: { success in
        if success {
            print("opened")
        } else {
            print("failed")
            // showInvalidUrlAlert()
        }
    })
}

Upvotes: 3

Moreira L
Moreira L

Reputation: 11

Inside of guard statement, you can throw an exception created by your application instead only put a return, like this:

guard let urlString = url, !urlString.isEmpty, let url = URL(string: urlString) else {
    throw ErrorEnum.invalidURL
}

With this approach, you can catch the error and send a UI Feedback for the User where it calls de func openURL.

Upvotes: 1

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