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.

CodePudding user response：

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;

CodePudding user response：

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.