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?)

该解决方案在可读性和可维护性之间取得了适当的平衡。

struct Card {

    // ...

    static func deck() -> Card[] {
        var deck = Card[]()
        for rank in Rank.Ace.toRaw()...Rank.King.toRaw() {
            for suit in [Suit.Spades, .Hearts, .Clubs, .Diamonds] {
                let card = Card(rank: Rank.fromRaw(rank)!, suit: suit)
                deck.append(card)
            }
        }
    return deck
    }
}

let deck = Card.deck()

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)")
}

你可以试着像这样列举

enum Planet: String {
    case Mercury
    case Venus
    case Earth
    case Mars

    static var enumerate: [Planet] {
        var a: [Planet] = []
        switch Planet.Mercury {
            case .Mercury: a.append(.Mercury); fallthrough
            case .Venus: a.append(.Venus); fallthrough
            case .Earth: a.append(.Earth); fallthrough
            case .Mars: a.append(.Mars)
        }
    return a
    }
}

Planet.enumerate // [Mercury, Venus, Earth, Mars]

有时，您可能会处理具有底层原始整数类型的枚举类型，这种类型在整个软件开发生命周期中都会发生变化。下面是一个很适合这种情况的例子:

public class MyClassThatLoadsTexturesEtc
{
    //...

    // Colors used for gems and sectors.
    public enum Color: Int
    {
        // Colors arranged in order of the spectrum.
        case First = 0
        case Red, Orange, Yellow, Green, Blue, Purple, Pink
        // --> Add more colors here, between the first and last markers.
        case Last
    }

    //...

    public func preloadGems()
    {
        // Preload all gems.
        for i in (Color.First.toRaw() + 1) ..< (Color.Last.toRaw())
        {
            let color = Color.fromRaw(i)!
            loadColoredTextures(forKey: color)
        }
    }

    //...
}

enum Rank: Int
{
    case Ace = 0
    case Two, Three, Four, Five, Six, Seve, Eight, Nine, Ten
    case Jack, Queen, King
    case Count
}

enum Suit : Int
{
    case Spades = 0
    case Hearts, Diamonds, Clubs
    case Count
}

struct Card
{
    var rank:Rank
    var suit:Suit
}

class Test
{
    func makeDeck() -> Card[]
    {
        let suitsCount:Int = Suit.Count.toRaw()
        let rankCount:Int = Rank.Count.toRaw()
        let repeatedCard:Card = Card(rank:Rank.Ace, suit:Suit.Spades)
        let deck:Card[] = Card[](count:suitsCount*rankCount, repeatedValue:repeatedCard)

        for i:Int in 0..rankCount
        {
            for j:Int in 0..suitsCount
            {
                deck[i*suitsCount+j] = Card(rank: Rank.fromRaw(i)!, suit: Suit.fromRaw(j)!)
            }
        }
        return deck
    }
}

根据Rick的回答:这要快5倍