Using JObject and JProperty with JSON.Net 4.0

Question

I'm trying to deserialize JSON in this format:

{
   "data": [
      {
         "installed": 1,
         "user_likes": 1,
         "user_education_history": 1,
         "friends_education_history": 1,
         "bookmarked": 1
      }
   ]
}

to a simple string array like this:

{
    "installed",
    "user_likes",
    "user_education_history",
    "friends_education_history",
    "bookmarked"
}

using JSON.NET 4.0

I've gotten it to work using the `CustomCreationConverter'

public class ListConverter : CustomCreationConverter<List<string>>
{
public override List<string> Create(Type objectType)
{
    return new List<string>();
}

public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
    var lst = new List<string>();

    //don't care about the inital 'data' element
    reader.Read();

    while (reader.Read())
    {
        if (reader.TokenType == JsonToken.PropertyName)
        {
            lst.Add(reader.Value.ToString());
        }
    }
    return lst;
}
}

but this really seems like overkill, especially if I want to create one for many different json responses.

I've tried using JObject but it doesn't seem like I'm doing it right:

List<string> lst = new List<string>();
JObject j = JObject.Parse(json_string);
foreach (JProperty p in j.SelectToken("data").Children().Children())
{
    lst.Add(p.Name);
}

Is there a better way to do this?

carlosfigueira · Accepted Answer · 2012-05-10 22:25:22Z

7

There are many ways you can do that, and what you have is fine. A few other alternatives are shown below:

Get the first element of the array, instead of all the children

Use SelectToken to go to the first array element with a single call

    string json = @"{
      ""data"": [
        {
          ""installed"": 1,
          ""user_likes"": 1,
          ""user_education_history"": 1,
          ""friends_education_history"": 1,
          ""bookmarked"": 1
        }
      ]
    }";

    JObject j = JObject.Parse(json);

    // Directly traversing the graph
    var lst = j["data"][0].Select(jp => ((JProperty)jp).Name).ToList();
    Console.WriteLine(string.Join("--", lst));

    // Using SelectToken
    lst = j.SelectToken("data[0]").Children<JProperty>().Select(p => p.Name).ToList();
    Console.WriteLine(string.Join("--", lst));

answered May 10, 2012 at 22:25

carlosfigueira

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2 Comments

Clox Over a year ago

Is there any difference between those 2 ways in performance or anything?

carlosfigueira Over a year ago

If any, I think will probably be negligible. But as always, try in your scenario, measure it, and decide for yourself.

Adrian Mole · Accepted Answer · 2020-10-13 14:17:18Z

1

You can use this solution as well to get the key and value:

string key, value;
var propertyList = (JObject)j["data"];
foreach (var property in propertyList)
{
   key = property.Key;
   value = property.Value.ToString();
}

edited Oct 13, 2020 at 14:17

Adrian Mole

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answered Oct 13, 2020 at 13:08

Muhammad Shoaib Karamat

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