swift - 类型为 'String' 的表达式模式无法匹配类型为“NSStoryboardSegue.Identifier”的值

Question

我正在尝试将我的 Swift 3 代码转换为 Swift 4。我收到此错误消息:

Expression pattern of type 'String' cannot match values of type 'NSStoryboardSegue.Identifier

这是我的代码:

override func prepare(for segue: NSStoryboardSegue, sender: Any?) {
    switch segue.identifier {

        case "showVC1":
            // DO SOMETHING
            break

        default:
            break
    }
}

我应该使用哪种类型而不是“String”？

Answer 1

从 Swift 4 开始， Storyboard标识符是一个可选的 NSStoryboardSegue.Identifier，定义为

extension NSStoryboardSegue {
    public struct Identifier : RawRepresentable, Equatable, Hashable {
        public init(_ rawValue: String)
        public init(rawValue: String)
    }
}

你可以打开它的rawValue:

    switch segue.identifier?.rawValue {
    case "showVC1"?:
        // do something ...
    default:
        break
    }

然而，推荐的模式是为每个定义常量 Storyboard标识符:

extension NSStoryboardSegue.Identifier {
    static let showVC1 = NSStoryboardSegue.Identifier("showVC1")
    // other storyboard identifiers ...
}

然后可以匹配:

    switch segue.identifier {
    case .showVC1?:
        // do something ...
    default:
        break
    }

在这两个示例中，“可选模式”x?(.some(x) 的快捷方式)用于匹配可选值。

为其他“标识符”引入了类似的类型，例如NSImage.Name，是NSImage(named:)的参数类型在 Swift 4 中。

有关更多信息，请参阅关于 swift-users 邮件的讨论列表，开始于

• [swift-users] What is up with names not being Strings any more in Swift 4? .

总体思路(据我了解)是创建单独的类型对于每种标识符。特别是(来自 https://lists.swift.org/pipermail/swift-users/Week-of-Mon-20170717/005940.html ):

... We are deliberately discouraging the string literal of the name. The string literal should be in only one place: the definition of the name constant. Everything else should use the constant. The compiler can provide autocompletion and typo detection of the constant. The string literal doesn't get that.