Question 111086 in Json Question posted in
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Trying to deserialize JSON returns null values

Using the method JsonConvert.DeserializeObject returns the default values for all properties.

var current = JsonConvert.DeserializeObject<Current>(myJson);

{
    "location": {
        "name": "London"
    },
    "current": {
        "temp_c": 5.0,
        "cloud": 50
    }
}

public class Current
{
    public double Temp_c { get; set; }
    public double Cloud { get; set; }
}

The expected current object should have the values: 50 for Cloud, and 5.0 for Temp_c, but returns the default values for all properties.

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2

Answers


  1. 0

    You need to define a class model like json object
    and then deserialize to it

    public class YourModel {

   //create location class that has Name property
   public Location Location { get; set; }

   //create current class that has Temp_c and Cloud property
   public Current Current { get; set; }

}

    and then

    var data = JsonConvert.DeserializeObject<YourModel>(myJson);

    and get the current value from data object

    var current = data.Current;
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  2. 0

    your ‘Current’ class is far of been like the JSON you post.

    You need to convert the JSON string to a C# class. You can use QuickType to convert it (Newtonsoft compatible).

    Note: I am using System.Text.Json.Serialization but the class model should be equal, just change:

    [JsonPropertyName("temp_c")] // .Net serializer (I prefer this)

    to

    [JsonProperty("temp_c")] // Newtonsoft.Json serializer

    (replace "temp_c" for every name)

    Here is the class model (a complete console application) you need:

    using System;
using System.Text.Json;
using System.Text.Json.Serialization;

#nullable disable

namespace test
{
    public class Weather
    {
        [JsonPropertyName("location")]
        public Location Location { get; set; }

        [JsonPropertyName("current")]
        public Current Current { get; set; }
    }

    public class Location
    {
        [JsonPropertyName("name")]
        public string Name { get; set; }
    }

    public class Current
    {
        [JsonPropertyName("temp_c")]
        public double TempC { get; set; }

        [JsonPropertyName("cloud")]
        public int Cloud { get; set; }
    }


    class Program
    {
        static void Main(string[] args)
        {
            string json = "{"location": { "name": "London" }, "current": { "temp_c": 5.0, "cloud": 50 }}";
            Weather myWeather = JsonSerializer.Deserialize<Weather>(json);

            Console.WriteLine("Location: {0} - Temp: {1:F}", myWeather.Location.Name, myWeather.Current.TempC);
        }
    }
}

    Now the Deserializer will work OK.

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