How to get element in JSON string

Question

I have this JSON string called assignee:

{
    "id": 15247055788906,
    "gid": "15247055788906",
    "name": "Bo Sundahl",
    "resource_type": "user"
}

I want to get the "name" element and its value if it's not null. I have tried

var jobject = JsonConvert.DeserializeObject<JObject>(assignee);

And

var jo = JObject.Parse(assignee);

I tried looping through it but I just get null exception or empty output even though if I just print the assignee variable itself its filled with data.

My loop is like:

foreach (var result in jobject["name"])
{
    Debug.WriteLine(result);
}

Parse it to a proper C# class instead of JObject, it makes all this complication go away. — DavidG
– DavidG, Commented Mar 1, 2019 at 12:58
its really not necessary for such small data, string name = jobject["name"]; was enough — user11110869
– user11110869, Commented Mar 1, 2019 at 13:03
There are many benefits though. Compile time type checking is the obvious one. However there are less tangible ones, for example the readability of the code. — DavidG
– DavidG, Commented Mar 1, 2019 at 13:04
@fasola it's not about the verbosity of the method, it's about the correct way to handle it. That's why I deleted my answer and upvoted david's — DotNetRussell
– DotNetRussell, Commented Mar 1, 2019 at 13:04
Another point to note is that deserialising to a concrete class is almost twice as fast as using JObject. — DavidG
– DavidG, Commented Mar 1, 2019 at 13:20

DavidG · Accepted Answer · 2019-03-01 13:13:42Z

3

The simplest and best way is to deserialise to a C# class, for example:

public class Data
{
    public long Id { get; set; }
    public string Name { get; set; }
    //etc..
}

And deserialise like this

var data = JsonConvert.DeserializeObject<Data>(json);
var name = data.Name;

edited Mar 1, 2019 at 13:13

answered Mar 1, 2019 at 13:03

DavidG

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4 Comments

Clay Over a year ago

Upvoted...but, if the incoming data is "dirty" and you don't know all the forms, this approach can be problematic.

DavidG Over a year ago

@Clay OP specifically needs the Name property, so that's not a concern here.

Clay Over a year ago

Agreed - just making a minor point. I do agree that this is the best approach...for both maintainability and performance reasons. I've experienced dirty data on larger json strings w/o knowing the source. Very unsatisfactory :-(

DotNetRussell Over a year ago

One work around for not knowing if a property exists or not is have a wrapper json package with meta data { "type" : "myType1", "Payload" : "{ inner json payload}" } This of course assumes you control the caller

Uwe Keim · Accepted Answer · 2019-03-01 13:01:46Z

1

To get name use this

string name = jobject["name"];

edited Mar 1, 2019 at 13:01

Uwe Keim

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answered Mar 1, 2019 at 12:57

Saif

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Comments

Youssef SABIH · Accepted Answer · 2019-03-01 13:27:03Z

0

Using ["name"] returns a JToken, it is null if the property doesn't exist

JToken token = jo["name"];
Debug.WriteLine(token?.ToString() ?? "<default value>");

answered Mar 1, 2019 at 13:27

Youssef SABIH

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Comments

ozanmut · Accepted Answer · 2019-03-01 14:31:37Z

0

If you don't know properties beforehand, you can loop through JObject properties and get name value pairs as following:

var jsonObject = JObject.Parse(str);
foreach (var item in jsonObject)
{
   var name = item.Key;
   JToken token = item.Value;
   if (token is JValue)
   {
      var value = token.Value<string>();
   }
}

answered Mar 1, 2019 at 14:31

ozanmut

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Comments

Dave · Accepted Answer · 2019-03-01 16:04:58Z

0

Here is how it should work:

class Data
  {
    public long? Id { get; set; }
    public string Gid { get; set; }
    public string Name { get; set; }
    public string Resource_Type { get; set; }
  }
  class Program
  {
    static void Main(string[] args)
    {

      string assignee = "{\"id\": 15247055788906, \"gid\": \"15247055788906\", \"name\": \"Bo Sundahl\", \"resource_type\": \"user\"}";
      Data data = JsonConvert.DeserializeObject<Data>(assignee);
      Console.WriteLine(data.Id);
      Console.WriteLine(data.Gid);
      Console.WriteLine(data.Name);
      Console.WriteLine(data.Resource_Type);
      Console.ReadLine();
    }
  }

answered Mar 1, 2019 at 16:04

Dave

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