这个问题现在简单多了。以下是我的Swift 4.2解决方案:

enum Suit: Int, CaseIterable {
  case None
  case Spade, Heart, Diamond, Club

  static let allNonNullCases = Suit.allCases[Spade.rawValue...]
}

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

  static let allNonNullCases = Rank.allCases[Two.rawValue...]
}

func makeDeck(withJoker: Bool = false) -> [Card] {
  var deck = [Card]()
  for suit in Suit.allNonNullCases {
    for rank in Rank.allNonNullCases {
      deck.append(Card(suit: suit, rank: rank))
    }
  }
  if withJoker {
    deck.append(Card(suit: .None, rank: .Joker))
  }
  return deck
}

4。2:

我喜欢这个解决方案，我把找到“列表理解在Swift”。

它使用Int rawws而不是string，但它避免了键入两次，它允许自定义范围，并且不硬编码原始值。

这是我最初解决方案的Swift 4版本，但请参阅上面的4.2改进:

enum Suit: Int {
  case None
  case Spade, Heart, Diamond, Club

  static let allRawValues = Suit.Spade.rawValue...Suit.Club.rawValue
  static let allCases = Array(allRawValues.map{ Suit(rawValue: $0)! })
}
enum Rank: Int {
  case Joker
  case Two, Three, Four, Five, Six
  case Seven, Eight, Nine, Ten
  case Jack, Queen, King, Ace

  static let allRawValues = Rank.Two.rawValue...Rank.Ace.rawValue
  static let allCases = Array(allRawValues.map{ Rank(rawValue: $0)! })
}
func makeDeck(withJoker: Bool = false) -> [Card] {
  var deck = [Card]()
  for suit in Suit.allCases {
    for rank in Rank.allCases {
      deck.append(Card(suit: suit, rank: rank))
    }
  }
  if withJoker {
    deck.append(Card(suit: .None, rank: .Joker))
  }
  return deck
}

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?)

如果您仍然想为Rank和Suit使用枚举，这里有一个不那么神秘的例子。如果您想使用for-in循环遍历每个对象，只需将它们收集到一个Array中。

标准52张牌的例子:

enum Rank: Int {
    case Ace = 1, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, Jack, Queen, King
    func name() -> String {
        switch self {
        case .Ace:
            return "ace"
        case .Jack:
            return "jack"
        case .Queen:
            return "queen"
        case .King:
            return "king"
        default:
            return String(self.toRaw())
        }
    }
}

enum Suit: Int {
    case Diamonds = 1, Clubs, Hearts, Spades
    func name() -> String {
        switch self {
        case .Diamonds:
            return "diamonds"
        case .Clubs:
            return "clubs"
        case .Hearts:
            return "hearts"
        case .Spades:
            return "spades"
        default:
            return "NOT A VALID SUIT"
        }
    }
}

let Ranks = [
    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 Suits = [
    Suit.Diamonds,
    Suit.Clubs,
    Suit.Hearts,
    Suit.Spades
]


class Card {
    var rank: Rank
    var suit: Suit

    init(rank: Rank, suit: Suit) {
        self.rank = rank
        self.suit = suit
    }
}

class Deck {
    var cards = Card[]()

    init() {
        for rank in Ranks {
            for suit in Suits {
                cards.append(Card(rank: rank, suit: suit))
            }
        }
    }
}

var myDeck = Deck()
myDeck.cards.count  // => 52

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)
}

更新代码:Swift 4.2/Swift 5

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

按问题访问输出:

for suitKey in Suit.allCases {
    print(suitKey.rawValue)
}

输出:

♠
♥
♦
♣

CaseIterable:提供其所有值的集合。 符合CaseIterable协议的类型通常是没有关联值的枚举。当使用CaseIterable类型时，您可以通过使用该类型的allCases属性访问该类型的所有案例的集合。

对于访问case，我们使用。allcases。更多信息请点击https://developer.apple.com/documentation/swift/caseiterable

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)")
}