从 Swift 2.2 更新后 switch 和 case 语句不起作用

Question

我正在尝试使用 Swift 制作一个股票应用程序，但我遇到了 switch 和 case 语句的问题。它给了我一个错误，它希望我在案例中添加一个问号，因此它将读取案例 ?1 而不是案例 1，当我这样做时，它给了我 3 个错误，说明预期模式、预期表达式和预期'：'在“案例”之后。任何帮助都会很棒。

func timeLabelsForTimeFrame(_ range: ChartTimeRange) -> [String] {

    switch range {
    case .oneDay:
        return ["9:30am", "10", "11", "12pm", "1", "2", "3", "4"]
    case .fiveDays:

        let weekday = (Calendar(identifier: 
Calendar.Identifier.gregorian) as NSCalendar).components(.weekday, 
from: Date()).weekday
        switch weekday {
        case 1:
        return ["Mon", "Tues", "Wed", "Thu", "Fri"]
        case 2:
        return ["Tues", "Wed", "Thu", "Fri", "Mon"]
        case 3:
        return ["Wed", "Thu", "Fri", "Mon", "Tues"]
        case 4:
        return ["Thu", "Fri", "Mon", "Tues", "Wed"]
        case 5:
        return ["Fri", "Mon", "Tues", "Wed", "Thu"]
        case 6:
        return ["Mon", "Tues", "Wed", "Thu", "Fri"]
        case 7:
        return ["Mon", "Tues", "Wed", "Thu", "Fri"]
        default: ()
        }
    case .tenDays:
        let weekday = (Calendar(identifier: Calendar.Identifier.gregorian) as NSCalendar).components(.weekday, from: Date()).weekday
        switch weekday {
        //sunday
        case 1:
        return ["Mon", "Wed", "Fri", "Mon", "Wed", "Fri"]
        case 2:
        return ["Wed", "Fri", "Mon", "Wed", "Fri", "Mon"]
        case 3:
        return ["Wed", "Fri", "Mon", "Wed", "Fri", "Tues"]
        case 4:
        return ["Fri", "Mon", "Wed", "Fri", "Mon", "Wed"]
        case 5:
        return ["Wed", "Mon", "Wed", "Fri", "Tues", "Thu"]
        case 6:
        return ["Mon", "Wed", "Fri", "Mon", "Wed", "Fri"]
        //saturday
        case 7:
        return ["Mon", "Wed", "Fri", "Mon", "Wed", "Fri"]
        default: ()
        }
    case .oneMonth:

        let fmt = DateFormatter()
        fmt.dateFormat = "dd MMM"
        let offset = Double(-6*24*60*60)
        let start = Date()
        let fifthString = fmt.string(from: start.addingTimeInterval(offset))
        let fourthString = fmt.string(from: start.addingTimeInterval(offset * 2))
        let thirdString = fmt.string(from: start.addingTimeInterval(offset * 3))
        let secondString = fmt.string(from: start.addingTimeInterval(offset * 4))
        let firstString = fmt.string(from: start.addingTimeInterval(offset * 5))

        return[firstString, secondString, thirdString, fourthString, fifthString, ""]
    case .threeMonths:
        let fmt = DateFormatter()
        fmt.dateFormat = "dd MMM"
        let offset = Double(-15*24*60*60)
        let start = Date()
        let fifthString = fmt.string(from: start.addingTimeInterval(offset))
        let fourthString = fmt.string(from: start.addingTimeInterval(offset * 2))
        let thirdString = fmt.string(from: start.addingTimeInterval(offset * 3))
        let secondString = fmt.string(from: start.addingTimeInterval(offset * 4))
        let firstString = fmt.string(from: start.addingTimeInterval(offset * 5))

        return[firstString, secondString, thirdString, fourthString, fifthString, ""]
    case .oneYear:
        let fmt = DateFormatter()
        fmt.dateFormat = "MMM"
        let offset = Double(-80*24*60*60)
        let start = Date()
        let fifthString = fmt.string(from: start.addingTimeInterval(offset))
        let fourthString = fmt.string(from: start.addingTimeInterval(offset * 2))
        let thirdString = fmt.string(from: start.addingTimeInterval(offset * 3))
        let secondString = fmt.string(from: start.addingTimeInterval(offset * 4))
        let firstString = fmt.string(from: start.addingTimeInterval(offset * 5))

        return[firstString, secondString, thirdString, fourthString, fifthString, ""]

    case .fiveYears:
        let fmt = DateFormatter()
        fmt.dateFormat = "yyyy"
        let offset = Double(-365*24*60*60)
        let start = Date()
        let fifthString = fmt.string(from: start.addingTimeInterval(offset))
        let fourthString = fmt.string(from: start.addingTimeInterval(offset * 2))
        let thirdString = fmt.string(from: start.addingTimeInterval(offset * 3))
        let secondString = fmt.string(from: start.addingTimeInterval(offset * 4))
        let firstString = fmt.string(from: start.addingTimeInterval(offset * 5))

        return[firstString, secondString, thirdString, fourthString, fifthString, ""]
    }
    return []

Answer 1

在 Swift 2 中，NSDateComponents 中的日期组件是 表示为Int，未定义的组件被设置为特殊值NSDateComponentUndefined。

在 Swift 3 中，DateComponents 中的日期组件表示为 可选项 (Int?) 和未定义的组件是 nil。

所以你的代码中的weekday 是可选的，你需要解开它 （这是安全的，因为您请求了该组件）：

switch weekday! {
case 1:
    // ...
}

或者你可以匹配“可选模式”：

switch weekday {
case 1?:
    // ...
}

（显然 Xcode "Fix-it" 不能正常工作，它提示 case ?1: 而不是 case 1?:。这发生在两个 Xcode 8 和 Xcode 9，看起来像一个错误。）

然而，一个更简单的解决方案是使用 component(_:from:) Calendar的方法：

let weekday = Calendar(identifier: .gregorian).component(.weekday, from: Date())

它为您提供单个日期组件作为（非可选）Int。 请注意，不需要转换为 NSCalendar。

Answer 2

这里发生错误是因为变量 weekday 是一个可选变量，而您在 switch case 语句中使用了它。您可以为weekday 提供默认值，也可以在使用 switch case 语句之前强制解包。

let weekday = (Calendar(identifier: Calendar.Identifier.gregorian) as NSCalendar).components(.weekday, from: Date()).weekday ?? 0

在这里，您将0 作为默认值提供给weekday。或者你可以使用

let weekday = (Calendar(identifier: Calendar.Identifier.gregorian) as NSCalendar).components(.weekday, from: Date()).weekday
   switch weekday! {
   case 1:
   .....
}