如何通过枚举值对对象进行排序？

Question

我有Workout类Difficulty财产

enum Difficulty: String { 
    case easy = "easy" 
    case moderate = "moderate" 
    case hard = "hard" 
} 

class Workout { 
    var name: String? 
    var difficulty: Difficulty? 
    . 
    . 
    . 
} 

我想用难度属性进行排序锻炼的数组。我知道我可以通过指定枚举的原始值，以int值实现这一点，如下比较这些值：

data.sort { $0.workout.difficulty!.rawValue < $1.workout.difficulty!.rawValue } 

但我真的希望这个枚举存储字符串，因为它的方便，它分配给标签文本上下行没有丑陋的开关箱黑客攻击，并且在某种程度上具有可比性。

如何实现这一目标？

Answer 1

你可以使用你的枚举散列值，以你的锻炼排序。您应该使用结构而不是类，并且类似于Igor所建议的结构，您可以使结构与枚举相媲美：

struct Workout: Comparable { 
    let name: String 
    let difficulty: Difficulty 
    enum Difficulty: String { 
     case easy, moderate, hard 
    } 
    static func ==(lhs: Workout, rhs: Workout) -> Bool { 
     return lhs.difficulty.hashValue == rhs.difficulty.hashValue 
    } 
    static func <(lhs: Workout, rhs: Workout) -> Bool { 
     return lhs.difficulty.hashValue < rhs.difficulty.hashValue 
    } 
} 

let wk1 = Workout(name: "night", difficulty: .hard) 
let wk2 = Workout(name: "morning", difficulty: .easy) 
let wk3 = Workout(name: "afternoon", difficulty: .moderate) 

let workouts = [wk1, wk2, wk3] // [{name "night", hard}, {name "morning", easy}, {name "afternoon", moderate}] 

let sorted = workouts.sorted() // [{name "morning", easy}, {name "afternoon", moderate}, {name "night", hard}] 

Answer 2

在您的enum上实施Comparable协议。它为您提供了一个static func < (lhs: Difficulty, rhs: Difficulty) -> Bool方法，您可以在其中定义排序。

下面是一个使用属性来简化整个样本订货

enum Difficulty: String, Comparable { 
    case easy = "easy" 
    case moderate = "moderate" 
    case hard(String) = "hard" 

    private var sortOrder: Int { 
     switch self { 
      case .easy: 
       return 0 
      case .moderate: 
       return 1 
      case .hard(_): 
       return 2 
     } 
    } 

    static func ==(lhs: Difficulty, rhs: Difficulty) -> Bool { 
     return lhs.sortOrder == rhs.sortOrder 
    } 

    static func <(lhs: Difficulty, rhs: Difficulty) -> Bool { 
     return lhs.sortOrder < rhs.sortOrder 
    } 
} 

使其能够使用

data.sort { $0.workout.difficulty! < $1.workout.difficulty! }