如何用$符号获取Json变量

Question

I've been pulling my hair to get the $type variable.

jsonTxt.json

{
  "$type": "Things.YourThings.YourThingName, Things",
  "Name": "Doe"      
}

I tried to get the variable as a string , but with no success. I just get null .

Here is what I do:

public class CustomName
{

  [JsonProperty("$type")]
  public string Type { get; set; }
  public string Name { get; set; }
}

Then,

var customName = JsonConvert.DeserializeObject<CustomName>(jsonText);

In actual fact, I just want to extract the type which is just the name YourThingName .

Answer 1

Try this:

JObject obj = JObject.Parse(jsonText);
var customName = new CustomName()
{
    Name = obj["Name"].ToString(),
    Type = obj["$type"].ToString()
};

Then to get just the YourThingName you can either use a Regex or just String.Split :

string name = Regex.Match(customName.Type, @"(?:\.)(\w*)(?:,)").Groups[1].ToString();

Or

string name = customName.Type.Split(',')[0].Split('.')[2];

You must do your boundary checks before accessing the different arrays or you'll end up with IndexOutOfRange exceptions.

.Net Fiddle

Answer 2

Another solution is to replace all occurences of "$type" with something like just "type" .

jsonText.Replace("\"$type\"", "\"type\"");

With...

public class CustomName
{
  public string Type { get; set; }
  public string Name { get; set; }
}

...deserialization will work as expected:

var customName = JsonConvert.DeserializeObject<CustomName>(jsonText);
var type = customName.Type;

Answer 3

Json.Net provides a MetadataPropertyHandling setting which controls how it handles $type , $ref and $id metadata properties in the JSON. By default it will consume these, meaning they are invisible to your classes. However, if you set this setting to Ignore , then Json.Net will not process metadata properties at all, allowing you to handle them normally. You don't need to resort to manipulating the JSON string manually beforehand.

string json = @"
{
  ""$type"": ""Things.YourThings.YourThingName, Things"",
  ""Name"": ""Doe""      
}";

JsonSerializerSettings settings = new JsonSerializerSettings
{
    MetadataPropertyHandling = MetadataPropertyHandling.Ignore
};

CustomName cn = JsonConvert.DeserializeObject<CustomName>(json, settings);

Console.WriteLine("Type: " + cn.Type);    // Things.YourThings.YourThingName, Things
Console.WriteLine("Name: " + cn.Name);    // Doe

From there you can extract the short type name like this:

int i = cn.Type.LastIndexOf(", ");
int j = cn.Type.LastIndexOf(".", i);
string shortTypeName = cn.Type.Substring(j + 1, i - j - 1);

Console.WriteLine(shortTypeName);    // YourThingName

Answer 4

There is another (stupid) solution to this problem. Create the same type as described in the $type property and deserialize to it:

// Put this in a library project called "Things"
namespace Things.YourThings
{
     public class YourThingName
     {
         public string Name { get; set; }
     }
}

Somewhere else:

var customName = JsonConvert.DeserializeObject<Things.YourThings.YourThingName>(jsonText);
var type = customName.GetType().Name;