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Json deserialize

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JohnKramer answered on June 14, 2020 Popularity 10/10 Helpfulness 2/10

answer Json deserialize

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Json deserialize

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List<Item> AllItems = new List<Item>();
            string json = File.ReadAllText("Items.json");
            AllItems = Deserialize<List<Item>>(json);
            WriteLine(AllItems[0].Name);

Popularity 10/10 Helpfulness 2/10 Language csharp

Source: stackoverflow.com

Tags: c# json

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Contributed on Jul 24 2022

JohnKramer

0 Answers Avg Quality 2/10

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.NET JSON: Deserialize

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using System.Text.Json;

namespace DeserializeFromFile
{
    public class WeatherForecast
    {
        public DateTimeOffset Date { get; set; }
        public int TemperatureCelsius { get; set; }
        public string? Summary { get; set; }
    }

    public class Program
    {
        public static void Main()
        {
            string fileName = "WeatherForecast.json";
            string jsonString = File.ReadAllText(fileName);
            WeatherForecast weatherForecast = JsonSerializer.Deserialize<WeatherForecast>(jsonString)!;

            Console.WriteLine($"Date: {weatherForecast.Date}");
            Console.WriteLine($"TemperatureCelsius: {weatherForecast.TemperatureCelsius}");
            Console.WriteLine($"Summary: {weatherForecast.Summary}");
        }
    }
}
// output:
//Date: 8/1/2019 12:00:00 AM -07:00
//TemperatureCelsius: 25
//Summary: Hot

Popularity 10/10 Helpfulness 4/10 Language csharp

Source: learn.microsoft.com

Tags: .net

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Contributed on Nov 25 2022

Arti

0 Answers Avg Quality 2/10

javascript json deserialize

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var objData = JSON.parse(json_string);

Popularity 10/10 Helpfulness 4/10 Language javascript

Source: Grepper

Tags: javascript j

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Contributed on Jun 14 2020

Worried Whale

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deserialize json c#

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using System;
using System.Text.Json;

public class Person
{
    public string Name { get; set; }
    public int Age { get; set; }
}

public class Program
{
    public static void Main()
    {
        string json = "{\"Name\":\"John\",\"Age\":30}";

        // Deserialize JSON to object
        Person person = JsonSerializer.Deserialize<Person>(json);

        Console.WriteLine("Name: " + person.Name);
        Console.WriteLine("Age: " + person.Age);
    }
}

Popularity 9/10 Helpfulness 4/10 Language csharp

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Contributed on Sep 11 2023

YOU.com

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java deserialize json object

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-1

@Test
public void whenDeserializingToSimpleObject_thenCorrect() {
    String json = "{"intValue":1,"stringValue":"one"}";

    Foo targetObject = new Gson().fromJson(json, Foo.class);

    assertEquals(targetObject.intValue, 1);
    assertEquals(targetObject.stringValue, "one");
}

Popularity 9/10 Helpfulness 1/10 Language java

Source: www.baeldung.com

Tags: java

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Contributed on Apr 17 2022

electriclance

0 Answers Avg Quality 2/10

Json deserialize

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.NET JSON: Deserialize

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