如何从 SWIFT 字典中检索深度嵌套的数据？

Question

我正在努力实现以下目标，需要您帮助我快速处理字典：

1. 如何获得“Nobody”“图像”值？它应该是“Nobody.png”
2. 如何计算 BestMoments 成员的数量？ （第一部电影应该是 3，第二部电影应该是 4"
3. 如何在 Bestmoments 中获取“Nobody”的值（例如：“Survive”、120、8）？

 var MV = [
        ["Nobody": [
            ["YEAR" : "2021"] ,
            ["IMAGE" : "Nobody.png"] ,
            ["GENRE" : "Action/Thriller"] ,
            ["DURATIONhours" : 1] ,
            ["DURATIONmintues" : 36] ,
            ["DURATIONseconds" : 17] ,
            ["BestMoments" : [["Win", 25, 5], ["Survive", 120, 8], ["Lose", 240, 15]]]
        ]],
        ["Godzilla vs. Kong": [
            ["YEAR" : "2021"] ,
            ["IMAGE" : "Nobody.png"] ,
            ["GENRE" : "Action/Sci-fi"] ,
            ["DURATIONhours" : 1] ,
            ["DURATIONmintues" : 22] ,
            ["DURATIONseconds" : 45] ,
            ["BestMoments" : [["Win", 34, 6], ["Survive", 120, 8], ["Twist", 233, 3], ["Lose", 340, 12]]]
        ]],
    ]

我尝试了以下方法来访问数据（带有结果）：

print(MV[0])
// 打印第一部电影的所有数据

print(MV["Nobody"])
// 对下标的调用没有完全匹配

print([MV[0].keys])
// 打印第一部电影的名字，它是键

for (movie, data) in MV2 {
        print(data)
        }

//元组模式不能匹配非元组类型的值 [String : [[String : Any]]]]

Answer 1

更好地表示数据：

struct BestMoment {
    enum MomentType: String {
        case win = "Win"
        case survive = "Survive"
        case lose = "Lose"
        case twist = "Twist"
    }

    let type: MomentType
    let start: Int
    let duration: Int
}

struct Duration {
    let hours: Int
    let minutes: Int
    let seconds: Int
}

struct Movie {
    let name: String
    let year: Int
    let image: String
    let genre: String
    let duration: Duration
    let bestMoments: [BestMoment]
}

let movies = [
    Movie(
        name: "Nobody",
        year: 2021,
        image: "Nobody.png",
        genre: "Action/Thriller",
        duration: Duration(hours: 1, minutes: 36, seconds: 16),
        bestMoments: [
            BestMoment(type: .win, start: 25, duration: 5),
            BestMoment(type: .survive, start: 120, duration: 8),
            BestMoment(type: .lose, start: 240, duration: 15)
        ]
    ),
    Movie(
        name: "Godzilla vs. Kong",
        year: 2021,
        image: "Nobody.png",
        genre: "Action/Sci-fi",
        duration: Duration(hours: 1, minutes: 22, seconds: 45),
        bestMoments: [
            BestMoment(type: .win, start: 34, duration: 6),
            BestMoment(type: .survive, start: 120, duration: 8),
            BestMoment(type: .twist, start: 233, duration: 3),
            BestMoment(type: .lose, start: 340, duration: 12)
        ]
    )
]

那么你的操作就简单了：

1. movies.first { $0.name == "Nobody" }?.image
2. movies.first { $0.name == "Nobody" }?.bestMoments.count
3. movies.first { $0.name == "Nobody" }?.bestMoments

您甚至可以将您的电影库包装到一个额外的对象中，以简化电影的搜索。

持续时间可能仅以秒表示，并且仅在需要时转换为小时/分钟/秒。 Genre 可能是 enum。

Answer 2

有问题的数据具有非常复杂的数据结构，难以管理。我认为有一个更简单的结构会更好。

但是，您可以通过以下方式访问所需的数据。 尝试打印结果。

/// 1. How to get "Nobody" "Image" value? it should be "Nobody.png"
let imageOfNobady = MV[0]["Nobody"]![1]["IMAGE"]

/// 2. How to the count of BestMoments members? (it should be 3 for first movie and 4 for second movie"
let bestmomentsForNobady = MV[0]["Nobody"]![6]["BestMoments"] as! [Any]
let countOfBestmomentsForNobady = bestmomentsForNobady.count

/// 3. how to get values inside Bestmoments for "Nobody" (like :"Survive", 120, 8)?
let survieInfo = bestmomentsForNobady[1]