enum Suit: String {
    case spades = "♠"
    case hearts = "♥"
    case diamonds = "♦"
    case clubs = "♣"
}

例如,我怎么做这样的事情:

for suit in Suit {
    // do something with suit
    print(suit.rawValue)
}

结果示例:

♠
♥
♦
♣

当前回答

对不起,我的回答是具体到我如何在我需要做的事情中使用这篇文章。对于那些无意中遇到这个问题的人,寻找一种方法在枚举中找到一个case,这是一种方法(Swift 2新增):

编辑:小写驼峰现在是Swift 3 enum值的标准

// From apple docs: If the raw-value type is specified as String and you don’t assign values to the cases explicitly, each unassigned case is implicitly assigned a string with the same text as the name of that case.

enum Theme: String
    {
    case white, blue, green, lavender, grey
    }

func loadTheme(theme: String)
    {
    // this checks the string against the raw value of each enum case (note that the check could result in a nil value, since it's an optional, which is why we introduce the if/let block
    if let testTheme = Theme(rawValue: theme)
        {
        // testTheme is guaranteed to have an enum value at this point
        self.someOtherFunction(testTheme)
        }
    }

对于那些对枚举感到疑惑的人来说,本页上给出的答案包括一个包含所有枚举值的数组的静态var/let是正确的。最新的苹果tvOS示例代码包含了完全相同的技术。

也就是说,他们应该在语言中构建一个更方便的机制(苹果,你在听吗?)

其他回答

我添加了函数count(),并迭代值:

public enum MetricType: Int {
    case mvps = 0
    case allNBA = 1
    case championshipRings = 2
    case finalAppearances = 3
    case gamesPlayed = 4
    case ppg = 5

    static func count() -> Int {
        return (ppg.rawValue) + 1
    }

    static var allValues: [MetricType] {
        var array: [MetricType] = Array()
        var item : MetricType = MetricType.mvps
        while item.rawValue < MetricType.count() {
            array.append(item)
            item = MetricType(rawValue: (item.rawValue + 1))!
        }
    return array
    }
}

在处理Swift 2.0时,以下是我的建议:

我已经将原始类型添加到Suit enum

enum Suit: Int {

然后:

struct Card {
    var rank: Rank
    var suit: Suit


    func fullDeck()-> [Card] {

        var deck = [Card]()

        for i in Rank.Ace.rawValue...Rank.King.rawValue {

            for j in Suit.Spades.rawValue...Suit.Clubs.rawValue {

                deck.append(Card(rank:Rank(rawValue: i)! , suit: Suit(rawValue: j)!))
            }
        }

        return deck
    }
}

我的解决方案是声明一个包含所有枚举可能性的数组。所以for循环可以遍历所有这些。

//Function inside struct Card
static func generateFullDeck() -> [Card] {
    let allRanks = [Rank.Ace, Rank.Two, Rank.Three, Rank.Four, Rank.Five, Rank.Six, Rank.Seven, Rank.Eight, Rank.Nine, Rank.Ten, Rank.Jack, Rank.Queen, Rank.King]
    let allSuits = [Suit.Hearts, Suit.Diamonds, Suit.Clubs, Suit.Spades]
    var myFullDeck: [Card] = []

    for myRank in allRanks {
        for mySuit in allSuits {
            myFullDeck.append(Card(rank: myRank, suit: mySuit))
        }
    }
    return myFullDeck
}

//actual use:
let aFullDeck = Card.generateFullDeck()    //Generate the desired full deck

var allDesc: [String] = []
for aCard in aFullDeck {
    println(aCard.simpleDescription())    //You'll see all the results in playground
}

(改进Karthik Kumar的回答)

这个解决方案是使用编译器来保证你不会错过一个case。

enum Suit: String {
    case spades = "♠"
    case hearts = "♥"
    case diamonds = "♦"
    case clubs = "♣"

    static var enumerate: [Suit] {
        switch Suit.spades {
        // make sure the two lines are identical ^_^
        case        .spades, .hearts, .diamonds, .clubs:
            return [.spades, .hearts, .diamonds, .clubs]
        }
    }
}

如果你给枚举一个原始的Int值,它将使循环更容易。

例如,你可以使用anyGenerator来获得一个可以枚举你的值的生成器:

enum Suit: Int, CustomStringConvertible {
    case Spades, Hearts, Diamonds, Clubs
    var description: String {
        switch self {
        case .Spades:   return "Spades"
        case .Hearts:   return "Hearts"
        case .Diamonds: return "Diamonds"
        case .Clubs:    return "Clubs"
        }
    }
    static func enumerate() -> AnyGenerator<Suit> {
        var nextIndex = Spades.rawValue
        return anyGenerator { Suit(rawValue: nextIndex++) }
    }
}
// You can now use it like this:
for suit in Suit.enumerate() {
    suit.description
}
// or like this:
let allSuits: [Suit] = Array(Suit.enumerate())

然而,这看起来像一个相当常见的模式,如果我们可以通过简单地遵循协议使任何枚举类型可枚举,这不是很好吗?有了Swift 2.0和协议扩展,现在我们可以了!

简单地添加到你的项目:

protocol EnumerableEnum {
    init?(rawValue: Int)
    static func firstValue() -> Int
}
extension EnumerableEnum {
    static func enumerate() -> AnyGenerator<Self> {
        var nextIndex = firstRawValue()
        return anyGenerator { Self(rawValue: nextIndex++) }
    }
    static func firstRawValue() -> Int { return 0 }
}

现在,任何时候你创建一个枚举(只要它有一个Int原始值),你可以通过遵循协议使它可枚举:

enum Rank: Int, EnumerableEnum {
    case Ace, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, Jack, Queen, King
}
// ...
for rank in Rank.enumerate() { ... }

如果你的枚举值不以0开头(默认值),重写firstRawValue方法:

enum DeckColor: Int, EnumerableEnum {
    case Red = 10, Blue, Black
    static func firstRawValue() -> Int { return Red.rawValue }
}
// ...
let colors = Array(DeckColor.enumerate())

最后一个Suit类,包括用更标准的CustomStringConvertible协议替换simpleDescription,看起来像这样:

enum Suit: Int, CustomStringConvertible, EnumerableEnum {
    case Spades, Hearts, Diamonds, Clubs
    var description: String {
        switch self {
        case .Spades:   return "Spades"
        case .Hearts:   return "Hearts"
        case .Diamonds: return "Diamonds"
        case .Clubs:    return "Clubs"
        }
    }
}
// ...
for suit in Suit.enumerate() {
    print(suit.description)
}

Swift 3语法:

protocol EnumerableEnum {
    init?(rawValue: Int)
    static func firstRawValue() -> Int
}

extension EnumerableEnum {
    static func enumerate() -> AnyIterator<Self> {
        var nextIndex = firstRawValue()

        let iterator: AnyIterator<Self> = AnyIterator {
            defer { nextIndex = nextIndex + 1 }
            return Self(rawValue: nextIndex)
        }

        return iterator
    }

    static func firstRawValue() -> Int {
        return 0
    }
}