类型为“字符串"的表达式模式不能与类型为"NSStoryboardSegue.Identifier"的值匹配 [英] Expression pattern of type 'String' cannot match values of type 'NSStoryboardSegue.Identifier

问题描述

我正在尝试将我的Swift 3代码转换为Swift4.我收到此错误消息:

I'm trying to convert my Swift 3 code to Swift 4. I get this error message:

字符串"类型的表达模式不能与"NSStoryboardSegue.Identifier"类型的值匹配

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
    }
}

我应使用哪种类型代替字符串"?

Which type should I use instead of "String"?

推荐答案

Swift 4将identifier属性的类型从String?切换为 RawRepresentable ，String的RawType.您可能需要将代码更改为if语句链，或显式使用rawValue:

Swift 4 switched the type of identifier property from String? to NSStoryboardSegue.Identifier?. The type is RawRepresentable, RawType of String. You may need to change your code to a chain of if statements, or use rawValue explicitly:

switch segue.identifier {
    case let x where x.rawValue == "showVC1":
        // DO SOMETHING
    break
    default:
    break
}

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