JSON 属性名称作为属性的整个 JSON 路径

Question

我希望每个 JSON 属性名称都代表它的路径。

public class Person
{
    public string FirstName { get; set; }
    public string LastName { get; set; }
    public Address Address { get; set; }
}

public class Address
{
    public string Street { get; set; }
    public string StreetNumber { get; set; }
}

使用 Newtonsoft.Json 我目前得到这个 JSON：

{
   "firstName":"Homer",
   "lastName":"Simpson",
   "address":{
      "street":"Evergreen Terrace",
      "streetNumber":"742"
   }
}

是否可以改为获取此 JSON：

{
   "person.firstName":"Homer",
   "person.lastName":"Simpson",
   "person.address":{
      "person.address.street":"Evergreen Terrace",
      "person.address.streetNumber":"742"
   }
}

Answer 1

你不能这样做，因为它是模棱两可的。这个文件如何反序列化？

在 C# 中，它可以被解释为...

public class Person
{
    public string Person.firstName { get; set; }
    public string Person.LastName { get; set; }
    public Address Person.Address { get; set; }
}

这是一种无效的格式，所以你的文件一定会导致异常或编译错误。

在浏览器上试试这个，使用JSON.parse() 方法，你会得到一个具有这种结构的对象（属性名称带有一个点）。

Answer 2

此格式无效。但 ： 样本

        private static string _getCustomJsonFormat(object o, string _parentPath = "")
    {
        string str = "{" ;
        {

            var t = o.GetType();
            var props = t.GetProperties();
            if (props != null)
            {
                List<string> _kv = new List<string>();
                foreach (var i in props)
                {
                    List<string> _path = new List<string>() { };
                    if (_parentPath.Length == 0) _path.Add(t.Name);
                    _path.AddRange(_parentPath.Split('.').Select(m => m.Trim()).Where(m => m.Length > 0));
                    string _p = i.Name; 
                    object _v = i.GetValue(o);
                    Type _vt = _v.GetType();
                    _path.Add(_p);
                    if (_vt.IsNested)
                    {
                        // Get Object
                        _kv.Add("\"" + string.Join(".",_path) + "\":" + _getCustomJsonFormat(_v, string.Join(".", _path)));
                    }
                    else {
                        // Get Property
                        _kv.Add("\"" + string.Join(".", _path)+ "\":\"" + _v.ToString() + "\"");
                    }
                }
                str += string.Join(",", _kv);

            }
        }
        return str+" }";
    }

使用：

            var p = new Person() { Address = new Address() { Street = "1", StreetNumber = "2" }, FirstName = "a", LastName = "b" };
        var x = _getCustomJsonFormat(p);

Answer 3

Json.NET 不会使用 JSONPaths 作为开箱即用的属性名称来序列化对象，因为它是基于协定的序列化程序。因此，它将以相同的方式序列化某种类型的所有实例T，无论它们在序列化图中遇到什么。相反，您希望为遇到的每个实例创建一个稍微不同的合约，以反映图中实例的路径。

那么，您有哪些选择来获取所需的 JSON？

首先，您可以序列化为JObject，然后使用public static void Rename(this JToken token, string newName) 扩展方法将其从this answer 由Brian Rogers 后处理到Rename JProperty in json.net：

var person = new Person
{
    // Initialize your Person as required
};

var settings = new JsonSerializerSettings
{
    ContractResolver = new CamelCasePropertyNamesContractResolver(),
};
var jObject = JObject.FromObject(person, JsonSerializer.CreateDefault(settings));
foreach (var item in jObject.Descendants().OfType<JProperty>().Select(p => (Property: p, p.Path)).ToList())
{
    // The Rename extension method from https://stackoverflow.com/a/47269811/3744182
    item.Property.Rename("person." + item.Path);
}

var json = jObject.ToString();

演示小提琴 #1 here.

其次，您可以在序列化时通过继承JsonTextWriter 并覆盖WritePropertyName() 方法来动态重命名属性：

public class NameRemappingJsonWriter : JsonTextWriter
{
    readonly Action<string> jsonWriterWritePropertyName;
    readonly Func<string, int, string, string> map;

    public NameRemappingJsonWriter(TextWriter textWriter, Func<string, int, string, string> map) : base(textWriter)
    {
        this.map = map ?? throw new ArgumentNullException(nameof(map));
        //Method to call a base-of-base-class method taken from this answer https://stackoverflow.com/a/32562464
        //By https://stackoverflow.com/users/5311735/evk
        //To https://stackoverflow.com/questions/2323401/how-to-call-base-base-method
        var ptr = typeof(JsonWriter).GetMethod("WritePropertyName", new[] { typeof(string) }).MethodHandle.GetFunctionPointer();
        jsonWriterWritePropertyName = (Action<string>)Activator.CreateInstance(typeof(Action<string>), this, ptr);
    }

    public override void WritePropertyName(string name) => WritePropertyName(name, true);

    public override void WritePropertyName(string name, bool escape)
    {
        jsonWriterWritePropertyName(name);
        WriteRaw(JsonConvert.ToString(map(name, Top, Path)));
        WriteRaw(":");
    }       
}

然后：

var settings = new JsonSerializerSettings
{
    ContractResolver = new CamelCasePropertyNamesContractResolver(),
};
Func<string, int, string, string> map = (name, depth, parentPath) => "person." + parentPath;
using var textWriter = new StringWriter();
using (var jsonWriter = new NameRemappingJsonWriter(textWriter, map) { Formatting = Formatting.Indented })
{
    JsonSerializer.CreateDefault(settings).Serialize(jsonWriter, person);
}
var json = textWriter.ToString();

请注意，要让它工作，我必须调用基类的基类 JsonWriter.WritePropertyName() 方法，这有点粗略和 not recommended or supported natively in c#。

演示小提琴 #2 here.