Convert JSON to C# POCO

Question

I have below JSON data

{
   "appDesc": {
                "description": "App description.",
                "message": "Create and edit presentations "
              },
   "appName": {
                "description": "App name.",
                "message": "Slides"
              }
}

I want to Deserialize into C# class object. I am using JsonConvert.DeserializeObject<>() to achieve this functionality. But some how it is not working.

 string JsonData= System.IO.File.ReadAllText(msgJSONpath);

 var moreInfo = JsonConvert.DeserializeObject<appName>(msg)



internal class appName
{
    public string message { get; set; }

    public string description { get; set; }
}

So moreInfo object will have 2 properties in message and description.

internal class appName { public string message { get; set; } public string description { get; set; } } — AMIT SHELKE, Nov 20 '18 at 06:50
Senario is, when I read a .json file which is having above data, I need to convert them into objects. Something like I have mentioned above. — AMIT SHELKE, Nov 20 '18 at 06:52
Possible duplicate of [How to Convert JSON object to Custom C# object?](https://stackoverflow.com/questions/2246694/how-to-convert-json-object-to-custom-c-sharp-object) — FarhadMohseni, Nov 20 '18 at 07:47

score 0 · Accepted Answer · answered Nov 20 '18 at 06:58

0

JObject defines method Parse for this:

JObject json = JObject.Parse(str);

or try for a typed object try:

Foo json  = JsonConvert.DeserializeObject<Foo>(str)

answered Nov 20 '18 at 06:58

A.M. Patel

334
2
9

JustLearning · Answer 2 · 2018-11-20T08:01:33.357

0

you need 2 C# classes since the properties of appName and appDesc are exactly the same.

To store appname

public class appName {
  public string description { get; set; }
  public string message { get; set; }
}

A class to have both above classes as properties

public class appResult {
  public appName appDesc { get; set; }
  public appName appName { get; set; }

  public appResult() {
    appDesc = new appName();
    appName = new appName();
  }
}
}

desirialize the json

var result = JsonConvert.DeserializeObject<appResult>(msg);

Once you have the result object you can get your appName
```
var appName = result.appName;
```

edited Nov 20 '18 at 08:01

answered Nov 20 '18 at 07:04

JustLearning

3,164
3
35
52

1

`appName` and `appDesc` are the same. There's no reason to use separate classes for them – Panagiotis Kanavos Nov 20 '18 at 07:50
@BackSlash i just saw it now, why it makes sense – JustLearning Nov 20 '18 at 08:00

FarhadMohseni · Answer 3 · 2018-11-20T08:01:47.697

First ,you need to create some classes based on your JSON , if you,re using visual studio you can copy the JSON string to your clipboard and than go to

Edit > Paste Special > Paste JSON As Classes

otherwise you can use This Online Tool

after that your code should be something like this :

     string JsonData= System.IO.File.ReadAllText(msgJSONpath);

     var moreInfo = JsonConvert.DeserializeObject<RootObject>(msg);

Generated Classes Based On Your JSON :

    public class AppDesc
{
    public string description { get; set; }
    public string message { get; set; }
}

public class AppName
{
    public string description { get; set; }
    public string message { get; set; }
}

public class RootObject
{
    public AppDesc appDesc { get; set; }
    public AppName appName { get; set; }
}

This would be a good answer if you pasted the generated classes as well — Panagiotis Kanavos, Nov 20 '18 at 07:50

Convert JSON to C# POCO

3 Answers3