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edited Sep 11, 2018 at 16:34

IfHow to install

Install-Package Newtonsoft.Json:

In Visual Studio, Tools menu -> Manage Nuget Package Manger Solution and type “JSON.NET” to search it online. Here's the figure:

Then create class with input parameters that you needexpect to get items from theit as a JSON object, I would use Jsonresponse.NET's LINQ

public class Inputs
{
    public string input1 { get; set; }
    public string input2 { get; set; }
    public string input3 { get; set; }
    public string input4 { get; set; }
}

Now we will convert it to JSON JObjecta .NET object using DeserializeObject() method of JsonConvert class. For exampleHere is the code:

JToken tokenprivate =void JObject.ParseJSONDeserilaize(stringFullOfJsonstring json);
int{
 page = (int)token.SelectToken("page");
int totalPagesvar obj = (int)tokenJsonConvert.SelectTokenDeserializeObject<Inputs>("total_pages"json);
}

I like this approach because you don't need to fully deserialize the JSON object. This comes in handy with APIs that can sometimes surpriseNow you with missinghave an object properties, like Twitter.

Documentation: Serializing and Deserializing JSON with Json.NET and LINQ to JSON with Json.NETof type Inputs Deserialized successfully

If you need to get items from the JSON object, I would use Json.NET's LINQ to JSON JObject class. For example:

JToken token = JObject.Parse(stringFullOfJson);
int page = (int)token.SelectToken("page");
int totalPages = (int)token.SelectToken("total_pages");

I like this approach because you don't need to fully deserialize the JSON object. This comes in handy with APIs that can sometimes surprise you with missing object properties, like Twitter.

Documentation: Serializing and Deserializing JSON with Json.NET and LINQ to JSON with Json.NET

How to install

Install-Package Newtonsoft.Json:

In Visual Studio, Tools menu -> Manage Nuget Package Manger Solution and type “JSON.NET” to search it online. Here's the figure:

Then create class with input parameters that you expect to get it as a JSON response.

public class Inputs
{
    public string input1 { get; set; }
    public string input2 { get; set; }
    public string input3 { get; set; }
    public string input4 { get; set; }
}

Now we will convert it to a .NET object using DeserializeObject() method of JsonConvert class. Here is the code:

private void JSONDeserilaize(string json)
{
    var obj = JsonConvert.DeserializeObject<Inputs>(json);
}

Now you have an object of type Inputs Deserialized successfully

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answered Sep 11, 2018 at 16:10

Di Kamal

If you need to get items from the JSON object, I would use Json.NET's LINQ to JSON JObject class. For example:

JToken token = JObject.Parse(stringFullOfJson);
int page = (int)token.SelectToken("page");
int totalPages = (int)token.SelectToken("total_pages");

I like this approach because you don't need to fully deserialize the JSON object. This comes in handy with APIs that can sometimes surprise you with missing object properties, like Twitter.

Documentation: Serializing and Deserializing JSON with Json.NET and LINQ to JSON with Json.NET

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