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
}

您可以通过实现ForwardIndexType协议来迭代枚举。

ForwardIndexType协议要求您定义一个继任者()函数来逐级遍历元素。

enum Rank: Int, ForwardIndexType {
    case Ace = 1
    case Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten
    case Jack, Queen, King

    // ... other functions

    // Option 1 - Figure it out by hand
    func successor() -> Rank {
        switch self {
            case .Ace:
              return .Two
            case .Two:
              return .Three

            // ... etc.

            default:
              return .King
        }
    }

    // Option 2 - Define an operator!
    func successor() -> Rank {
        return self + 1
    }
}

// NOTE: The operator is defined OUTSIDE the class
func + (left: Rank, right: Int) -> Rank {
    // I'm using to/from raw here, but again, you can use a case statement
    // or whatever else you can think of

    return left == .King ? .King : Rank(rawValue: left.rawValue + right)!
}

在开或闭范围内迭代(..<或…)将在内部调用继任者()函数，允许你这样写:

// Under the covers, successor(Rank.King) and successor(Rank.Ace) are called to establish limits
for r in Rank.Ace...Rank.King {
    // Do something useful
}

(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]
        }
    }
}

在Swift中，枚举类型可以像EnumType一样访问。案例:

let tableView = UITableView(frame: self.view. view)UITableViewStyle.Plain)

大多数情况下，只有当您有几个选项可以使用，并且确切地知道在每个选项上要做什么时，才会使用枚举类型。

在处理枚举类型时，使用for-in结构没有太大意义。

你可以这样做，例如:

func sumNumbers(numbers : Int...) -> Int {
    var sum = 0

    for number in numbers{
        sum += number
    }

    return sum
}

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?)

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)
}