使用路径和值更新 JSON 对象

Question

我有一个csv文件，其中包含以下格式的路径和值：

path;value
prop1.prop2.1;hello
prop1.prop2.2;world
prop1.prop2.3;!
prop1.prop3.test;hi
prop1.prop4;value

我魔杖把它当作json：

{
  "prop1": {
    "prop2": {
      "1": "hello",
      "2": "world",
      "3": "!"
    }
    "prop3": {
      "test": "hi"
    }
    "prop4": "value"
  }    
}

我已经解析了这样的csv文件：

Dictionary<string, string> dict = new Dictionary<string, string>();
while (csv.Read())
{
  string path = csv.GetField<string>(0);
  string value = csv.GetField<string>(1);
  dict.Add(path, value);
}

你能帮我提供方法吗，它将使用 JSON.Net 库从这个字典中创建 JSON。当然，原始文件中的属性可以不同。

Answer 1

您可以使用此函数从字典中将 JSON 作为字符串返回

public static string BuildJson(Dictionary<string, string> dict)
{
    dynamic expando = new ExpandoObject();
    foreach (KeyValuePair<string, string> pair in dict.Where(x => !string.Equals(x.Key, "path")))
    {
        string[] pathArray = pair.Key.Split('.');
        var currentExpando = expando as IDictionary<string, Object>;
        for (int i = 0; i < pathArray.Count(); i++)
        {
            if (i == pathArray.Count() - 1)
            {
                currentExpando.Add(pathArray[i], pair.Value);
            }
            else
            {
                if (!currentExpando.Keys.Contains(pathArray[i]))
                {
                    currentExpando.Add(pathArray[i], new ExpandoObject());
                }
                currentExpando = currentExpando[pathArray[i]] as IDictionary<string, Object>;
            }
        }
    }
    JObject o = JObject.FromObject(expando);
    return o.ToString();
}

您需要使用系统动态添加;

Answer 2

您可以利用 System.Dynamic.ExpandoObject：

public string ToJson(Dictionary<string, string> props)
{
    IDictionary<string, object> json = new System.Dynamic.ExpandoObject() as IDictionary<string, object>;
    foreach (var prop in props)
    {
        string path = prop.Key;
        if (String.IsNullOrWhiteSpace(path)) continue;
        string[] keys = path.Split('.');
        string value = prop.Value;
        var cursor = json;
        for (int i = 0; i < keys.Length; i++)
        {
            object innerJson;
            if (!cursor.TryGetValue(keys[i], out innerJson))
                cursor.Add(keys[i], new System.Dynamic.ExpandoObject() as IDictionary<string, object>);
            if (i == keys.Length - 1)
                cursor[keys[i]] = value;
            cursor = cursor[keys[i]] as IDictionary<string, object>;
        }
    }
    return JsonConvert.SerializeObject(json);
}

Answer 3

不使用动态：

Dictionary<string, object> dictionary = new Dictionary<string, object>();
while (csv.Read())
{
  string path = csv.GetField<string>(0);
  string value = csv.GetField<string>(1);
  dictionary = GetHierarchy(path, value);
}
string serialized = JsonConvert.SerializeObject(dictionary);
Console.WriteLine(serialized);

添加以下方法：

public Dictionary<string, object> GetHierarchy(Dictionary<string, object> root, string path, string value)
{
    string[] parts = path.Split('.');
    if (parts.Length > 1)
    {
        if(!root.ContainsKey(parts[0]))
            root.Add(parts[0], new Dictionary<string, object>());
        Dictionary<string, object> dict = root[parts[0]] as Dictionary<string, object>;
        GetMe(dict, path.Replace(parts[0] + ".", string.Empty), value);
    }
    else 
        root[parts[0]] = value;
    return root;
}

生成预期的输出