How to Use JsonSerializer in C# for JSON Conversion

Use JsonSerializer.Serialize() to convert C# objects to JSON strings and JsonSerializer.Deserialize() to convert JSON strings back to objects. These methods are part of System.Text.Json namespace and provide fast, easy JSON handling.

Syntax

The JsonSerializer class provides two main methods:

• Serialize(object value): Converts a C# object to a JSON string.
• Deserialize<T>(string json): Converts a JSON string back to a C# object of type T.

Both methods are static and part of the System.Text.Json namespace.

csharp

string jsonString = JsonSerializer.Serialize(objectToConvert);
MyClass obj = JsonSerializer.Deserialize<MyClass>(jsonString);

Example

This example shows how to serialize a simple Person object to JSON and then deserialize the JSON back to a Person object.

csharp

using System;
using System.Text.Json;

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

class Program
{
    static void Main()
    {
        Person person = new Person { Name = "Alice", Age = 30 };

        // Convert object to JSON string
        string jsonString = JsonSerializer.Serialize(person);
        Console.WriteLine("Serialized JSON: " + jsonString);

        // Convert JSON string back to object
        Person deserializedPerson = JsonSerializer.Deserialize<Person>(jsonString);
        Console.WriteLine($"Deserialized Person: Name={deserializedPerson.Name}, Age={deserializedPerson.Age}");
    }
}

Output

Serialized JSON: {"Name":"Alice","Age":30}
Deserialized Person: Name=Alice, Age=30

Common Pitfalls

Common mistakes when using JsonSerializer include:

• Not including using System.Text.Json; which causes methods to be unavailable.
• Trying to deserialize JSON into a type that does not match the JSON structure, causing exceptions.
• Ignoring case sensitivity: property names in JSON must match C# property names by default (case-sensitive).
• Not handling null values or missing properties which can cause errors or unexpected results.

To fix case sensitivity, use JsonSerializerOptions with PropertyNameCaseInsensitive = true.

csharp

using System.Text.Json;

var options = new JsonSerializerOptions
{
    PropertyNameCaseInsensitive = true
};

string json = "{\"name\":\"Bob\", \"age\":25}";
Person p = JsonSerializer.Deserialize<Person>(json, options);

Quick Reference

Here is a quick summary of key JsonSerializer methods and options:

Method/Option	Description
Serialize(object value)	Converts an object to a JSON string.
Deserialize(string json)	Converts JSON string to an object of type T.
JsonSerializerOptions.PropertyNameCaseInsensitive	Allows case-insensitive matching of JSON property names.
JsonSerializerOptions.WriteIndented	Formats JSON output with indentation for readability.

Key Takeaways

Use JsonSerializer.Serialize() to convert objects to JSON strings easily.

Use JsonSerializer.Deserialize() to parse JSON strings back to objects safely.

Include System.Text.Json namespace to access JsonSerializer methods.

Handle case sensitivity with JsonSerializerOptions to avoid deserialization errors.

Use options like WriteIndented for readable JSON output.