如何在Swift中创建嵌套的字典元素?

Fat*_*ebi 1 dictionary swift

我想创建一个存储此变量:

["messageCode": API_200, "data": {
    activities =     (
                {
            action = 1;
            state = 1;
        }
    );
    messages =     (
                {
            body = hi;
            // ...
        }
    );
}, "message": ]
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我所做的是这样的:

var fullDict: Dictionary<String, AnyObject> = [:]
fullDict["messageCode"] = "API_200" as AnyObject

var data: Dictionary<String, AnyObject> = [:]
fullDict ["data"] = data as AnyObject
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这样是否正确,我如何添加活动?

Ahm*_*d F 5

我建议创建一个自定义模型:

struct Model {
    var messageCode: String
    var data: MyData
    var message: String
}

struct MyData {
    let activities: [Activity]
    let messages: [Message]
}

struct Activity {
    var action: Int
    var state: Int
}

struct Message {
    var body: String
    // ...
}
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因此,您可以将其用作:

let data = MyData(activities: [Activity(action: 1, state: 1)], messages: [Message(body: "hi")])
let myModel = Model(messageCode: "API_200", data: data, message: "")
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但是,如果出于某种原因必须将其声明为字典,则可能是这样的:

let myDict: [String: Any] = [
                            "messageCode": "API_200",
                             "data": ["activities": [["action": 1, "state": 1]],
                                      "messages": [["body": "hi"]]
                                     ],
                             "message": ""
                            ]
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这意味着该myDict词典包含:

  • messageCode 串。
  • data作为嵌套字典,其中包含:
    • activities字典数组(的数组[String: Int])。
    • messages字典数组(的数组[String: String])。
  • message 串。

您应该采用建模方法的最简单原因之一是,当要从中读取时myModel,您要做的就是使用点.符号。与将其用作字典不同,您必须区分其值,这在某些时候可能令人头疼。例如,假设我们要访问数据消息数组中的第一个消息正文:

模型:

myModel.data.messages.first?.body
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字典:

if let data = myDict["data"] as? [String: [[String: Any]]],
   let messages = data["messages"] as? [[String: String]],
   let body = messages.first?["body"] {

    print(body)
}
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