How do you handle a JSON response in Java?

Handling JSON in Java can be done using Jackson, Gson, or org.json.

• Jackson (best for large projects): Use ObjectMapper to parse JSON into Java objects.

  ObjectMapper objectMapper = new ObjectMapper();
  User user = objectMapper.readValue(json, User.class);
  

• Gson (lightweight): Use Gson to convert JSON to objects.

  Gson gson = new Gson();
  User user = gson.fromJson(json, User.class);
  

• org.json (manual handling): Parse JSON using JSONObject.

  JSONObject obj = new JSONObject(json);
  String name = obj.getString("name");
  

Jackson is the most powerful choice.

Handling a JSON response in Java depends on the library you’re using. Here is most common way:

1. Using Jackson (Most Popular)

Jackson is a widely used library for handling JSON in Java.

To use it, add this dependency (if using Maven):

<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-databind</artifactId>
<version>2.15.0</version>
</dependency>

Example: Convert JSON to Java Object (Deserialization)

import com.fasterxml.jackson.databind.ObjectMapper;

public class Main {
public static void main(String[] args) throws Exception {
String jsonResponse = "{ \"name\": \"John\", \"age\": 30 }";

// Convert JSON string to Java Object
ObjectMapper objectMapper = new ObjectMapper();
Person person = objectMapper.readValue(jsonResponse, Person.class);

System.out.println(person.getName() + " is " + person.getAge() + " years old.");
}
}

// Java class to map JSON data
class Person {
private String name;
private int age;

// Getters and Setters (Required for Jackson)
public String getName() { return name; }
public void setName(String name) { this.name = name; }

public int getAge() { return age; }
public void setAge(int age) { this.age = age; }
}