c# 解析轻松的 json 以制作一棵树

问题描述

所以我需要解析看起来像这样的文件:

So I need to parse file looking alike this:

{
pl: {
    GENERIC: {
        BACK: "COFNIJ",
        WAIT: "CZEKAJ",

        PAGES: {
            ABOUTME: {
                ID: "ID",
            },
            INFO: {
                STATUS: "STATUS",
            }
        }
    },

    TOP_MENU: {
        LOGGED: "Zalogowany",
        OPTIONS: "Opcje",
    }
},

en: {
    GENERIC: {
        BACK: "BACK",
        WAIT: "WAIT",

        PAGES: {
            ABOUTME: {
                ID: "ID",
            },
            INFO: {
                STATUS: "STATUS",
            }
        }
    },

    TOP_MENU: {
        LOGGED: "Logged",
        OPTIONS: "Options",
    }
}

}

但是我不知道这个文件有多少个元素，所以我无法创建类来解析这个文件.

But I don't know how many elements the file will have, so I can't create class to parse this file.

1. 我的第一个问题是如何在 C# 中用双引号将文件中的无引号"元素包裹起来，使该文件 json 可解析?
2. 如何将上面的 json 文件解析为树状数据结构，使其看起来像这样:样本树，所以我可以在控制台上输出每一片叶子带有路径的节点，en"和pl"子树中的值?
   例如:路径:generic/back en:"back" pl:"cofnij".

我已经尝试过使用 Dictionary字典 = JsonConvert.DeserializeObject在将上述宽松的 json 转换为有效的 json 之后，获取主键，但我认为树结构将是更有效的方法.感谢您的帮助！

I've already tried to use Dictionary<string, dynamic> dictionary = JsonConvert.DeserializeObject<Dictionary<string, dynamic>>(file);to get main keys, after converting above relaxed json to valid json, but I think the tree structure will be the more effective way.Thanks for any help!

推荐答案

您的第一个问题已经在这里问过了:解析非标准JSON

Your first question was already asked here: Parsing non-standard JSON

第二个问题听起来有点像这样:您的第一个问题已经在这里问过了:将 JSON 反序列化为 C# 动态对象?

The second question sounds a bit like this one: Your first question was already asked here: Deserialize JSON into C# dynamic object?

你可以创建一个动态对象

You could create a dynamic object

   dynamic myObj =   JsonConvert.DeserializeObject(json);
   foreach (Newtonsoft.Json.Linq.JProperty jproperty in myObj)
   {
         //..
   }

然后对其进行预处理以创建树结构.这可能会有所帮助:我如何反思动态对象?

and then preprocess it to create the tree structure. This could help: How do I reflect over the members of dynamic object?

这是通过遍历属性将反序列化动态转换为树结构的方法:

This is how you can convert your deserialized dynamic to a tree structure by iterating through the properties:

    public void Convert()
    {
        dynamic myObj = JsonConvert.DeserializeObject(json);
        PrintObject(myObj, 0);
    }

    private void PrintObject(JToken token, int depth)
    {
        if (token is JProperty)
        {
            var jProp = (JProperty)token;
            var spacer = string.Join("", Enumerable.Range(0, depth).Select(_ => "	"));
            var val = jProp.Value is JValue ? ((JValue)jProp.Value).Value : "-";

            Console.WriteLine($"{spacer}{jProp.Name}  -> {val}");

            foreach (var child in jProp.Children())
            {
                PrintObject(child, depth + 1);
            }
        }
        else if (token is JObject)
        {
            foreach (var child in ((JObject)token).Children())
            {
                PrintObject(child, depth + 1);
            }
        }
    }

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