Deserialization of a json string gives null value

Question

{
  "apple": {
    "A": "xyz",
    "B": "abc",
    "C": "jkl"
  },
  "banana": {
    "A": "lotus",
    "B": "oil",
    "C": "cat"
  }
}

This is my JSON and below is my model class where I want to map the JSON data.

 public class Wrapper
    {
        public Dictionary<string, item> fruits{ get; set; }
    }
public class item
    {
        public string A{get; set;}
        public string B{get; set;}
        public string C{get; set;}
    }

when I am using the following code to deserialize the Json string I am getting null as response.

 var value=Newtonsoft.Json.JsonConvert.DeserializeObject<Wrapper>(jsonString);

Your wrapper type expects an object with a single property, fruits. Your JSON doesn't define an object with a fruits property. Your JSON would have to be { "fruits": { "apple": {/*...*/}, "banana": {/*...*/} }. — T.J. Crowder
– T.J. Crowder, Commented Feb 17, 2022 at 12:00

Serge · Accepted Answer · 2022-02-17 12:00:17Z

3

you don't need any wrapper

var value=Newtonsoft.Json.JsonConvert.DeserializeObject<Dictionary<string, item>>(jsonString);

answered Feb 17, 2022 at 12:00

Serge

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Comments

David McEleney · Accepted Answer · 2022-02-17 12:02:17Z

1

Try wrap in an element fruits:

{
  "fruits":
  {
    "apple": {
      "A": "xyz",
      "B": "abc",
      "C": "jkl"
    },
    "banana": {
      "A": "lotus",
      "B": "oil",
      "C": "cat"
    }
  } 
}

To validate that your input is correct - instantiate an instance of the wrapper class and serialise it - then you can compare that your input matches the structure.

edited Feb 17, 2022 at 12:02

answered Feb 17, 2022 at 12:01

David McEleney

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3 Comments

T.J. Crowder Over a year ago

Why not just deserialize what they have, as Serge shows?

David McEleney Over a year ago

Because that doesn't match what's being asked in the question.

T.J. Crowder Over a year ago

We read the question different, but that's okay.

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