Question 85678 in ASP.NET Question posted in
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Assign Json to a string without serilization in c# – Asp.net

I have a JSON string

{
  "F1":
  { 
    "BF1":"BV1",
    "BF2":"BV2"
  },
  "F2":"BF1"
}

and I wish to assign it to an object of this type

public class MyType
{
   public string F1 {get;set;} 
   public string F2 {get;set;}
}

in other words, I need to convert JSON to an object, but the inner object is to be assigned as a JSON string

Tags:

2

Answers


  1. 0

    the simpliest way is

        var jObj = JObject.Parse(json);

    MyType myType = new MyType
    {
        F1 = jObj["F1"].ToString(),
        F2 = jObj["F2"].ToString()
    };

    or if you don’t want indented formating

    MyType myType = new MyType
    {
        F1 = JsonConvert.SerializeObject(jObj["F1"]),
        F2 = (string) jObj["F2"];
    };

    of if you like a hard way

    MyType myType = JsonConvert.DeserializeObject<MyType>(json,new MyTypeJsonConverter());

public class MyTypeJsonConverter : JsonConverter<MyType>
{
    public override MyType ReadJson(JsonReader reader, Type objectType, MyType existingValue, bool hasExistingValue, JsonSerializer serializer)
    {
        var jObj = JObject.Load(reader);

        return new MyType
        {
            F1 = JsonConvert.SerializeObject(jObj["F1"]),
            F2 = (string)jObj["F2"]
        };
    }

    // or maybe this 
    // public override bool CanWrite => false;

    public override void WriteJson(JsonWriter writer, MyType value, JsonSerializer serializer)
    {
        var val = new
        {
            F1 = JObject.Parse(value.F1),
            F2 = value.F2
        };
        
        writer.WriteRaw(JsonConvert.SerializeObject(val, Newtonsoft.Json.Formatting.Indented));
    }
    
}

    or if you have many properties that should deserialize common way

    MyType myType = JsonConvert.DeserializeObject<MyType>(json);

public class MyType
{ 
   [JsonConverter(typeof(MyTypePropertyJsonConverter))]
    public string F1 { get; set; }
    
    public string F2 { get; set; }
}

public class MyTypePropertyJsonConverter : JsonConverter<string>
{
    public override string ReadJson(JsonReader reader, Type objectType, string existingValue, bool hasExistingValue, JsonSerializer serializer)
    {
        return JsonConvert.SerializeObject(JToken.Load(reader));
    }

    public override void WriteJson(JsonWriter writer, string value, JsonSerializer serializer)
    {
        writer.WriteRawValue(value);
    }
}
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  2. 0

    Though the answer posted by @serge looks best to me. But I spent some time to resolve your issue and I ended up with the below code:

    dynamic jsonElement = JsonSerializer.Deserialize<dynamic>(json);
MyType myType = new MyType();
myType.F1 = jsonElement.GetProperty("F1").ToString();
myType.F2 = jsonElement.GetProperty("F2").ToString();       
Console.WriteLine(myType.F1 + ", " + myType.F2);

    I hope it may help you some way.

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