Question

我正在尝试随机化arrayList，然后将URL拉到IBAction，然后打开页面。

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]
}

下面是IBAction但我收到一条错误，说pickArrayList [2]是一个未解析的标识符。

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

提前致谢， Swift Noobie

Answer 1

发生错误是因为pickArrayList在viewDidLoad范围内被声明为局部变量。

将pickArrayList声明为类顶层的属性

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]
}

请遵守变量名称以小写字母开头的命名约定。

从Array中拉出URL字符串并放在IBAction中

1 个答案: