ios - 可选类型 'UIWindow?' 的值必须解包才能引用包装基类型 'viewWithTag' 的成员 'UIWindow'

Question

我已将 Objective-C 文件转换为 Swift，现在转换为:

if isCaptured {
    let colourView = UIView(frame: window.frame)
    colourView?.backgroundColor = UIColor.black
    colourView?.tag = 1234
    colourView?.alpha = 0

    window!.makeKeyAndVisible()
    window.addSubview(colourView)

    // fade in the view
    UIView.animate(withDuration: 0.5, animations: {
        colourView?.alpha = 1
    })
} else {
    // grab a reference to our coloured view
    let colourView = window.viewWithTag(1234)

    // fade away colour view from main view
    UIView.animate(withDuration: 0.5, animations: {
        colourView?.alpha = 0
    }) { finished in
        // remove when finished fading
        colourView?.removeFromSuperview()
    }
}

但我收到错误:

Value of optional type 'UIWindow?' must be unwrapped to refer to member 'viewWithTag' of wrapped base type 'UIWindow'

附上截图:

如何纠正这个问题？

Answer 1

您已经使用 window!.makeKeyAndVisible() 将其打开。因此，您可以在这里执行相同的操作 -> window!.viewWithTag(1234)。

还建议阅读https://docs.swift.org/swift-book/LanguageGuide/TheBasics.html 。您需要了解基本概念才能了解强制展开意味着什么以及何时可以安全使用。