Get Required Property from Complex json string in C# with Newtonsoft

Introduction

In this blog, I will explain the trick to convert JSON string property to C# Object using Newtonsoft.

Illustration

Suppose the JSON string is like…

{
  "Success": true,
  "ComplexObject": {
    "NormalProp": "Test",
    "ListOfAnotherClass": [{
      "JustAStringProp": "We"
    }, {
      "JustAStringProp": "Got"
    }, {
      "JustAStringProp": "Values"
    }]
  }
}

Now, if I get this string from some method and try to process, I would need to convert this to C# objects. The structure would look something like below. To easily get the class structure, you can use any online tool like json2csharp.

class Result
{
    public bool Success { get; set; }
    public ComplextJsonClass ComplexObject { get; set; }
}

class ComplextJsonClass
{
    public string NormalProp { get; set; }
    public List<AnotherClass> ListOfAnotherClass { get; set; }
}

class AnotherClass
{
    public string JustAStringProp { get; set; }
}

What We Need?

So, for processing, I would like to get the “ComplexObject” property of this JSON string converted as “ComplextJsonClass” C# object. As the JSON string is complex with other properties like “Success“, we only need the property “ComplexObject“, yet converted to a C# object.

How To Do That?

We will use Newtonsoft. You can easily find that from Nuget and add that as a reference to your project.

Solution

First, I will tell you how to generate this type of JSON.

var complex = new ComplextJsonClass
{
    NormalProp = "Test",
    ListOfAnotherClass = new List<AnotherClass>
    {
        new AnotherClass{JustAStringProp = "We"},
        new AnotherClass{JustAStringProp = "Got"},
        new AnotherClass{JustAStringProp = "Values"},
    }
};

var moreComplex = new Result {Success = true, ComplexObject = complex};
var jsonString = JsonConvert.SerializeObject(moreComplex);

This “jsonString” comes as I mentioned at the beginning of the post. Now, we need to convert this JSON to a C# Object. For this, we need to use JsonConvert.DeserializeObject Method (String).

    var jsonObject = JsonConvert.DeserializeObject(jsonString);

But, unfortunately, this alone won’t help us more to access the properties. We have to convert to a JObject, by which we can easily access the properties from the JSON.

var jsonObject = (JObject)JsonConvert.DeserializeObject(jsonString);

After this, it is just a matter of getting the property by using jsonObject["ComplexObject"].

var jsonObject = (JObject)JsonConvert.DeserializeObject(jsonString);
if (jsonObject["ComplexObject"] == null) return;

ComplextJsonClass complexObject = JsonConvert.DeserializeObject<ComplextJsonClass>
            (jsonObject["ComplexObject"].ToString());

Full Code

var complex = new ComplextJsonClass
{
    NormalProp = "Test",
    ListOfAnotherClass = new List<AnotherClass>
    {
        new AnotherClass{JustAStringProp = "We"},
        new AnotherClass{JustAStringProp = "Got"},
        new AnotherClass{JustAStringProp = "Values"},
    }
};

// Get the jsonString from complex object.
var moreComplex = new Result {Success = true, ComplexObject = complex};
var jsonString = JsonConvert.SerializeObject(moreComplex);

//Now let's deserialize and convert to our required ComplextJsonClass object.
var jsonObject = (JObject)JsonConvert.DeserializeObject(jsonString);
if (jsonObject["ComplexObject"] == null) return;

ComplextJsonClass complexObject = JsonConvert.DeserializeObject<ComplextJsonClass>
            (jsonObject["ComplexObject"].ToString());

Feedback

Let me know if this trick helped you in any way in your projects or assignments. Re-post or share, if you liked it.

Thanks a lot for reading. Merry Christmas and Happy New Year. 😀🤝