How Does JsonConvert.DeserializeObject() Work in C#?

Question

Khushi Singh · Answer

In C#, JsonConvert.DeserializeObject() turns text that follows JSON format into managed .NET objects. The
JsonConvert.DeserializeObject() method exists within Newtonsoft.Json library which people employ for JSON and
.NET object operations in their C# programming work. This tool lets developers convert JSON data from all sources into matched programming objects.
The main task of JsonConvert.DeserializeObject() is to recreate complex object networks from JSON text records for smooth data transfer across systems. This system works with basic and advanced object types at the same time for lists and dictionaries. To use this method developers must indicate what kind of object is targeted to convert the JSON data properly.
The following example explains its usage:
using Newtonsoft.Json;
using System;
class Program
{
   static void Main()
   {
       string jsonString = "{ \"Name\": \"John Doe\", \"Age\": 30, \"Email\": \"john@example.com\" }";
       Person person = JsonConvert.DeserializeObjectandlt;Personandgt;(jsonString);
       Console.WriteLine($"Name: {person.Name}, Age: {person.Age}, Email: {person.Email}");
   }
}
class Person
{
   public string Name { get; set; }
   public int Age { get; set; }
   public string Email { get; set; }
}
Output:
Name: John Doe, Age: 30, Email: john@example.com
This approach helps JSON data match correctly with C# objects which makes it simpler to handle formatted data in applications. The method handles API reactions and reads configuration files while saving data in JSON format.
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How Does JsonConvert.DeserializeObject() Work in C#?

Ashutosh Kumar Verma

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