如何在Swift中创建嵌套的字典元素？ [英] How to create nested dictionary elements in Swift?

问题描述

我想创建一个存储该变量的变量：

I want to create a variable which stores this:

["messageCode": API_200, "data": {
    activities =     (
                {
            action = 1;
            state = 1;
        }
    );
    messages =     (
                {
            body = hi;
            // ...
        }
    );
}, "message": ]

我所做的是这样：

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

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

是这种方式是正确的，我如何添加活动？

Is this way is correct and how I can add activities?

推荐答案

因为您明确希望将其作为 [String：AnyObject ] ：

Since you explicitly want it as [String:AnyObject]:

var dict: [String:AnyObject] = ["messageCode":"API_200" as AnyObject,
                                "data": ["activities": [["action":1,
                                                         "state":1]],
                                         "messages": [["body":"hi"]]] as AnyObject,
                                "message": "" as AnyObject]

基本上所有根值都应按类型转换为AnyObject

Basically all the root values should be typecasted as AnyObject

或者很长距离：

//Activities is as Array of dictionary with Int values
var activities = [[String:Int]]()
activities.append(["action": 1,
                   "state": 1])

//Messages is an Array of string
var messages = [[String:String]]()
messages.append(["body" : "hi"])

//Data is dictionary containing activities and messages
var data = [String:Any]()
data["activities"] = activities
data["messages"] = messages

//Finally your base dictionary
var dict = [String:AnyObject]()
dict["messageCode"] = "API_200" as AnyObject
dict["data"] = data as AnyObject
dict["message"] = "" as AnyObject
print(dict)

解析为获取您的数据将是地狱；

示例（让我们捕获动作）：

let action = ((dict["data"] as? [String:Any])?["activities"] as? [String:Int])?.first?.value

如您所见，您需要在每个级别进行转换。这是在Swift中使用字典的问题。

As you can see you need to typecast at every level. This is the problem with using dictionaries in Swift. Too much cruft.

当然，您可以使用第三方库，例如 SwiftyJSON 可以将上述内容简化为：

Sure, you could use a third-party library like SwiftyJSON to reduce the above to:

let action = dict["data"]["activities"][0]["action"]

但是您是否需要依赖项

如果定义了结构，则创建模型；正如 Ahmad F的答案所建议的那样。

If your structure is defined then create models instead; as Ahmad F's answer suggests. It will be more readable, maintainable and flexible.

...但是您可能会问，这就是纯 Dictionary 的方式。 code>元素。

...but since you asked, this is how one would do it with pure Dictionary elements.

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