这是一个相当老的帖子，来自Swift 2.0。现在有一些更好的解决方案，使用了swift 3.0的新特性: 在Swift 3.0中迭代一个Enum

关于这个问题，有一个解决方案，它使用了Swift 4.2的一个新功能(在我写这篇编辑时还没有发布): 我如何得到一个Swift枚举的计数?

在这个帖子中有很多好的解决方案，但其中一些非常复杂。我喜欢尽可能地简化。这里有一个解决方案，可能适用于不同的需求，但我认为它在大多数情况下都很好:

enum Number: String {
    case One
    case Two
    case Three
    case Four
    case EndIndex

    func nextCase () -> Number
    {
        switch self {
        case .One:
            return .Two
        case .Two:
            return .Three
        case .Three:
            return .Four
        case .Four:
            return .EndIndex

        /* 
        Add all additional cases above
        */
        case .EndIndex:
            return .EndIndex
        }
    }

    static var allValues: [String] {
        var array: [String] = Array()
        var number = Number.One

        while number != Number.EndIndex {
            array.append(number.rawValue)
            number = number.nextCase()
        }
        return array
    }
}

迭代:

for item in Number.allValues {
    print("number is: \(item)")
}

这篇文章是相关的https://www.swift-studies.com/blog/2014/6/10/enumerating-enums-in-swift

基本上，提议的解决方案是

enum ProductCategory : String {
     case Washers = "washers", Dryers = "dryers", Toasters = "toasters"

     static let allValues = [Washers, Dryers, Toasters]
}

for category in ProductCategory.allValues{
     //Do something
}

它花了我一点，而不仅仅是一个方法在结构像swift书调用，但我在枚举中设置了下一个函数。我会使用一个协议，我不知道为什么，但有秩设置为int混乱。

enum Rank: Int {
    case Ace = 1
    case Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten
    case Jack, Queen, King
    func simpleDescription() -> String {
        switch self {
        case .Ace:
            return "ace"
        case .Jack:
            return "jack"
        case .Queen:
            return "Queen"
        case .King:
            return "King"
        default:
            return String(self.toRaw())
        }
    }
    mutating func next() -> Rank {
        var rank = self
        var rawrank = rank.toRaw()
        var nrank: Rank = self
        rawrank = rawrank + 1
        if let newRank = Rank.fromRaw(rawrank) {
            println("\(newRank.simpleDescription())")
            nrank = newRank
        } else {
            return self
        }
        return nrank
    }
}

enum Suit {
    case Spades, Hearts, Diamonds, Clubs
    func color() -> String {
        switch self {
        case .Spades, .Clubs:
            return "black"
        default:
            return "red"
        }
    }
    func simpleDescription() -> String {
        switch self {
        case .Spades:
            return "spades"
        case .Hearts:
            return "hearts"
        case .Diamonds:
            return "diamonds"
        case .Clubs:
            return "clubs"
        }
    }
    mutating func next() -> Suit {
        switch self {
        case .Spades:
            return Hearts
        case .Hearts:
            return Diamonds
        case .Diamonds:
            return Clubs
        case .Clubs:
            return Spades
        }
    }
}

struct Card {
    var rank: Rank
    var suit: Suit
    func deck() -> Card[] {
        var tRank = self.rank
        var tSuit = self.suit
        let tcards = 52 // we start from 0
        var cards: Card[] = []
        for i in 0..tcards {
            var card = Card(rank: tRank, suit: tSuit)
            cards.append(card)
            tRank = tRank.next()
            tSuit = tSuit.next()
        }
        return cards
    }
    func simpleDescription() -> String {
        return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
    }
}

var card = Card(rank: .Ace, suit: .Spades)
var deck = card.deck()

我使用了一些常识，但这可以通过将花色乘以等级来轻松纠正(如果你没有使用标准的桥牌，你必须相应地改变枚举，如果基本上只是通过不同的枚举进行步骤)。为了节省时间，我使用了ranks rawValues，如果你愿意，你也可以为西装做同样的事情。然而，这个例子没有它，所以我决定在不改变suit rawValue的情况下找出它

另一个解决方案:

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

    static var count: Int {
        return 4   
    }

    init(index: Int) {
        switch index {
            case 0: self = .spades
            case 1: self = .hearts
            case 2: self = .diamonds
            default: self = .clubs
        }
    }
}

for i in 0..<Suit.count {
    print(Suit(index: i).rawValue)
}

更新到Swift 2.2+

func iterateEnum<T: Hashable>(_: T.Type) -> AnyGenerator<T> {
    var i = 0
    return AnyGenerator {
        let next = withUnsafePointer(&i) {
            UnsafePointer<T>($0).memory
        }
        if next.hashValue == i {
            i += 1
            return next
        } else {
            return nil
        }
    }
}

它更新了Swift 2.2表单@Kametrixom的答案

Swift 3.0+(非常感谢@Philip)

func iterateEnum<T: Hashable>(_: T.Type) -> AnyIterator<T> {
    var i = 0
    return AnyIterator {
        let next = withUnsafePointer(&i) {
            UnsafePointer<T>($0).pointee
        }
        if next.hashValue == i {
            i += 1
            return next
        } else {
            return nil
        }
    }
}

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
}