Convert JSON to C# POCO











up vote
-1
down vote

favorite












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.










share|improve this question




















  • 1




    show your class you want to de-serialize into
    – Just code
    Nov 20 at 6:49










  • internal class appName { public string message { get; set; } public string description { get; set; } }
    – AMIT SHELKE
    Nov 20 at 6:50










  • edit your question and add your relevant code
    – Just code
    Nov 20 at 6:51










  • 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 at 6:52










  • Add your code which you have tried
    – Rahul Neekhra
    Nov 20 at 6:55















up vote
-1
down vote

favorite












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.










share|improve this question




















  • 1




    show your class you want to de-serialize into
    – Just code
    Nov 20 at 6:49










  • internal class appName { public string message { get; set; } public string description { get; set; } }
    – AMIT SHELKE
    Nov 20 at 6:50










  • edit your question and add your relevant code
    – Just code
    Nov 20 at 6:51










  • 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 at 6:52










  • Add your code which you have tried
    – Rahul Neekhra
    Nov 20 at 6:55













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











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.










share|improve this 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.







c# json






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 20 at 7:58









JustLearning

94921634




94921634










asked Nov 20 at 6:48









AMIT SHELKE

2023923




2023923








  • 1




    show your class you want to de-serialize into
    – Just code
    Nov 20 at 6:49










  • internal class appName { public string message { get; set; } public string description { get; set; } }
    – AMIT SHELKE
    Nov 20 at 6:50










  • edit your question and add your relevant code
    – Just code
    Nov 20 at 6:51










  • 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 at 6:52










  • Add your code which you have tried
    – Rahul Neekhra
    Nov 20 at 6:55














  • 1




    show your class you want to de-serialize into
    – Just code
    Nov 20 at 6:49










  • internal class appName { public string message { get; set; } public string description { get; set; } }
    – AMIT SHELKE
    Nov 20 at 6:50










  • edit your question and add your relevant code
    – Just code
    Nov 20 at 6:51










  • 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 at 6:52










  • Add your code which you have tried
    – Rahul Neekhra
    Nov 20 at 6:55








1




1




show your class you want to de-serialize into
– Just code
Nov 20 at 6:49




show your class you want to de-serialize into
– Just code
Nov 20 at 6:49












internal class appName { public string message { get; set; } public string description { get; set; } }
– AMIT SHELKE
Nov 20 at 6:50




internal class appName { public string message { get; set; } public string description { get; set; } }
– AMIT SHELKE
Nov 20 at 6:50












edit your question and add your relevant code
– Just code
Nov 20 at 6:51




edit your question and add your relevant code
– Just code
Nov 20 at 6:51












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 at 6:52




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 at 6:52












Add your code which you have tried
– Rahul Neekhra
Nov 20 at 6:55




Add your code which you have tried
– Rahul Neekhra
Nov 20 at 6:55












3 Answers
3






active

oldest

votes

















up vote
0
down vote



accepted










JObject defines method Parse for this:



JObject json = JObject.Parse(str);


or try for a typed object try:



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





share|improve this answer




























    up vote
    0
    down vote













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





    1. To store appname



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



    2. 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();
      }
      }
      }



    3. desirialize the json



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



    4. Once you have the result object you can get your appName



      var appName = result.appName;







    share|improve this answer



















    • 1




      appName and appDesc are the same. There's no reason to use separate classes for them
      – Panagiotis Kanavos
      Nov 20 at 7:50










    • i see your point your are right
      – JustLearning
      Nov 20 at 7:59










    • @BackSlash i just saw it now, why it makes sense
      – JustLearning
      Nov 20 at 8:00


















    up vote
    0
    down vote













    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; }
    }





    share|improve this answer



















    • 1




      This would be a good answer if you pasted the generated classes as well
      – Panagiotis Kanavos
      Nov 20 at 7:50











    Your Answer






    StackExchange.ifUsing("editor", function () {
    StackExchange.using("externalEditor", function () {
    StackExchange.using("snippets", function () {
    StackExchange.snippets.init();
    });
    });
    }, "code-snippets");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "1"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53387635%2fconvert-json-to-c-sharp-poco%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    JObject defines method Parse for this:



    JObject json = JObject.Parse(str);


    or try for a typed object try:



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





    share|improve this answer

























      up vote
      0
      down vote



      accepted










      JObject defines method Parse for this:



      JObject json = JObject.Parse(str);


      or try for a typed object try:



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





      share|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        JObject defines method Parse for this:



        JObject json = JObject.Parse(str);


        or try for a typed object try:



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





        share|improve this answer












        JObject defines method Parse for this:



        JObject json = JObject.Parse(str);


        or try for a typed object try:



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






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 20 at 6:58









        user2156791

        628




        628
























            up vote
            0
            down vote













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





            1. To store appname



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



            2. 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();
              }
              }
              }



            3. desirialize the json



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



            4. Once you have the result object you can get your appName



              var appName = result.appName;







            share|improve this answer



















            • 1




              appName and appDesc are the same. There's no reason to use separate classes for them
              – Panagiotis Kanavos
              Nov 20 at 7:50










            • i see your point your are right
              – JustLearning
              Nov 20 at 7:59










            • @BackSlash i just saw it now, why it makes sense
              – JustLearning
              Nov 20 at 8:00















            up vote
            0
            down vote













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





            1. To store appname



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



            2. 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();
              }
              }
              }



            3. desirialize the json



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



            4. Once you have the result object you can get your appName



              var appName = result.appName;







            share|improve this answer



















            • 1




              appName and appDesc are the same. There's no reason to use separate classes for them
              – Panagiotis Kanavos
              Nov 20 at 7:50










            • i see your point your are right
              – JustLearning
              Nov 20 at 7:59










            • @BackSlash i just saw it now, why it makes sense
              – JustLearning
              Nov 20 at 8:00













            up vote
            0
            down vote










            up vote
            0
            down vote









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





            1. To store appname



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



            2. 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();
              }
              }
              }



            3. desirialize the json



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



            4. Once you have the result object you can get your appName



              var appName = result.appName;







            share|improve this answer














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





            1. To store appname



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



            2. 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();
              }
              }
              }



            3. desirialize the json



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



            4. Once you have the result object you can get your appName



              var appName = result.appName;








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 20 at 8:01

























            answered Nov 20 at 7:04









            JustLearning

            94921634




            94921634








            • 1




              appName and appDesc are the same. There's no reason to use separate classes for them
              – Panagiotis Kanavos
              Nov 20 at 7:50










            • i see your point your are right
              – JustLearning
              Nov 20 at 7:59










            • @BackSlash i just saw it now, why it makes sense
              – JustLearning
              Nov 20 at 8:00














            • 1




              appName and appDesc are the same. There's no reason to use separate classes for them
              – Panagiotis Kanavos
              Nov 20 at 7:50










            • i see your point your are right
              – JustLearning
              Nov 20 at 7:59










            • @BackSlash i just saw it now, why it makes sense
              – JustLearning
              Nov 20 at 8:00








            1




            1




            appName and appDesc are the same. There's no reason to use separate classes for them
            – Panagiotis Kanavos
            Nov 20 at 7:50




            appName and appDesc are the same. There's no reason to use separate classes for them
            – Panagiotis Kanavos
            Nov 20 at 7:50












            i see your point your are right
            – JustLearning
            Nov 20 at 7:59




            i see your point your are right
            – JustLearning
            Nov 20 at 7:59












            @BackSlash i just saw it now, why it makes sense
            – JustLearning
            Nov 20 at 8:00




            @BackSlash i just saw it now, why it makes sense
            – JustLearning
            Nov 20 at 8:00










            up vote
            0
            down vote













            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; }
            }





            share|improve this answer



















            • 1




              This would be a good answer if you pasted the generated classes as well
              – Panagiotis Kanavos
              Nov 20 at 7:50















            up vote
            0
            down vote













            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; }
            }





            share|improve this answer



















            • 1




              This would be a good answer if you pasted the generated classes as well
              – Panagiotis Kanavos
              Nov 20 at 7:50













            up vote
            0
            down vote










            up vote
            0
            down vote









            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; }
            }





            share|improve this answer














            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; }
            }






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 20 at 8:01

























            answered Nov 20 at 7:43









            FarhadMohseni

            215




            215








            • 1




              This would be a good answer if you pasted the generated classes as well
              – Panagiotis Kanavos
              Nov 20 at 7:50














            • 1




              This would be a good answer if you pasted the generated classes as well
              – Panagiotis Kanavos
              Nov 20 at 7:50








            1




            1




            This would be a good answer if you pasted the generated classes as well
            – Panagiotis Kanavos
            Nov 20 at 7:50




            This would be a good answer if you pasted the generated classes as well
            – Panagiotis Kanavos
            Nov 20 at 7:50


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Stack Overflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53387635%2fconvert-json-to-c-sharp-poco%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            To store a contact into the json file from server.js file using a class in NodeJS

            Redirect URL with Chrome Remote Debugging Android Devices

            Dieringhausen