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Deserialize the following in c#

Posted on 2014-04-08
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Last Modified: 2014-04-26
I am new to deserialization using JSON, so I need to know how to create a class object that gets assigned the values from the deserialization process.  I have the following data retrieved in JSON that I would like to properly create a class and be able to throw that information into the created class.  Please help!  Thanks!

{
    "response":
     {
                  "authToken":  "webapi123431095"
     },
    "status":
     {
              "detail":
               {
               },
               "success": true
     }
}

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Question by:VBBRett
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4 Comments
 
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Assisted Solution

by:plusone3055
plusone3055 earned 250 total points
ID: 39986799
I think this had already been addresses in a previous question :)

http://www.experts-exchange.com/Programming/Languages/C_Sharp/Q_28268323.html
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Author Comment

by:VBBRett
ID: 39986867
Thanks, but since I am new to deserializing, do you know of a way to deserialize the string that I posted?  Thanks!
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LVL 75

Expert Comment

by:käµfm³d 👽
ID: 39987463
There are plenty of examples on how to do this. For instance, my comment here:

http://www.experts-exchange.com/Programming/Languages/.NET/Q_28400081.html#a39962468

The basic idea is that anywhere in your JSON that you have curly braces ( {} ), you will be creating a new class; anywhere in your JSON that you have square brackets ( [] ), you will be creating an array; everything else is just a property. You need to create a "root" class to hold the overall JSON once it's converted.
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Accepted Solution

by:
käµfm³d   👽 earned 250 total points
ID: 39987472
In your case, you might define a root class as:

public class ServiceResponse
{
}

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...which will have two properties:

public class ServiceResponseModel
{
    public XXXX Response { get; set; }
    pulbic YYYY Status { get; set; }
}

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Now, both response and status are followed by curly braces, so that means you need a new class for each:

public class ResponseModel
{
    public string AuthToken { get; set; }
}

public class StatusModel
{
    ZZZZ  // I'll let you fill in ZZZZ  = )
}

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These new classes will be the types for the two new properties:

public class ServiceResponseModel
{
    public ResponseModel Response { get; set; }
    pulbic StatusModel Status { get; set; }
}

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Once you fill in everything, it's just a matter of passing the data and the type information (i.e. ServiceResponseModel) to the serializer.
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