Pull URL String from Array and place in IBAction

Question

I am trying to randomize the arrayList and then pull the URL to an IBAction that would then open the page.

override func viewDidLoad() {
    let ArrayList = [
        ["A", "A2", "http://a.com"],
        ["B", "B2", "http://b.com"],
        ["C", "C2", "http://c.com"],
    ]

    let pickArrayList = ArrayList[Int(arc4random_uniform(UInt32(ArrayList.count)))]
    label1.text = pickArrayList[0]
    label2.text = pickArrayList[1]
}

Below is the IBAction but I receive an error saying that pickArrayList[2] is an unresolved Identifier.

@IBAction func didTapButton(sender: AnyObject) {
        UIApplication.shared.open(URL(pickArrayList[2])!)  
    }

Thanks in advance, Swift Noobie

vadian · Accepted Answer · 2017-11-18 19:35:26Z

1

The error occurs because pickArrayList is declared as local variable in the scope of viewDidLoad.

Declare pickArrayList as a property on the top level of the class

var pickArrayList : [String]!

override func viewDidLoad() {
    super.viewDidLoad()
    let arrayList = [
        ["A", "A2", "http://a.com"],
        ["B", "B2", "http://b.com"],
        ["C", "C2", "http://c.com"],
    ]

    pickArrayList = arrayList[Int(arc4random_uniform(UInt32(arrayList.count)))]
    label1.text = pickArrayList[0]
    label2.text = pickArrayList[1]
}

Please conform to the naming convention that variable names start with lowercase letter.

answered Nov 18, 2017 at 19:35

vadian

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4 Comments

Chris Over a year ago

Man this is so close. I get a "fatal error" because it says there is a bad execution within the @IBAction. "UIApplication.shared.open(URL(string: pickArrayList[2])!)" If I remove the "string:" then I get an error that "Argument Labels (_:) do not match any available overloads."

vadian Over a year ago

URL(string: .. is mandatory. Make sure that string in pickArrayList[2] is a valid URL format or use optional binding.

Chris Over a year ago

Okay, so I have tried a bunch of different URLs and none have worked thus far. When I change the string value to "google.com" as opposed to pickArrayList[2], it has no problem. Any thoughts?

vadian Over a year ago

Once again, the string must be convertible to URL. Check the url in a browser or a playground. URL(string: returns nil if the URL format is not valid. But this is beyond the issue of the question.

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