Message serialization in Spring Boot - Mini Project: Build & Apply

Message serialization

📖 Scenario: You are building a simple Spring Boot application that sends messages between components. To do this, you need to convert a Java object into a JSON string format. This process is called message serialization.Imagine you want to send a greeting message from one part of your app to another. You will create a message object, configure the serialization, convert the object to JSON, and then complete the setup so the message can be sent.

🎯 Goal: Build a Spring Boot application that serializes a Message object into a JSON string using Jackson library.

📋 What You'll Learn

Create a Message class with a text field

Create a message object with the text 'Hello, Spring!'

Use ObjectMapper to serialize the message object to JSON

Print the JSON string to verify serialization

💡 Why This Matters

🌍 Real World

Message serialization is used in web services and APIs to send data between client and server in a format both can understand.

💼 Career

Understanding serialization is essential for backend developers working with REST APIs, microservices, and messaging systems.

1

Create the Message class

Create a public class called Message with a private String field called text. Add a public constructor that takes a String parameter text and assigns it to the field. Also add a public getter method called getText() that returns the text field.

Spring Boot

// Your code here

Hint

Think of Message as a simple box holding your text. You need a way to put text in (constructor) and take it out (getter).

2

Create a Message object

Create a variable called message of type Message and assign it a new Message object with the text "Hello, Spring!".

Spring Boot

public class Message {
    private String text;

    public Message(String text) {
        this.text = text;
    }

    public String getText() {
        return text;
    }
}

// Your code here

Hint

Think of message as your box filled with the greeting text.

3

Serialize the Message object to JSON

Create an ObjectMapper instance called mapper. Use mapper.writeValueAsString(message) to convert the message object to a JSON string and assign it to a variable called jsonString.

Spring Boot

public class Message {
    private String text;

    public Message(String text) {
        this.text = text;
    }

    public String getText() {
        return text;
    }
}

Message message = new Message("Hello, Spring!");

// Your code here

Hint

Use Jackson's ObjectMapper to turn your message box into a JSON string.

4

Print the JSON string

Add a line to print the jsonString variable using System.out.println(jsonString); to verify the serialization result.

Spring Boot

import com.fasterxml.jackson.databind.ObjectMapper;

public class Message {
    private String text;

    public Message(String text) {
        this.text = text;
    }

    public String getText() {
        return text;
    }
}

Message message = new Message("Hello, Spring!");

ObjectMapper mapper = new ObjectMapper();
String jsonString = mapper.writeValueAsString(message);

// Your code here

Hint

Print the JSON string to see the message in JSON format.

Practice

1. What is the main purpose of message serialization in Spring Boot?

easy

A. To handle user authentication and authorization

B. To compile Java code into bytecode

C. To convert objects into a format like JSON for easy data sharing

D. To manage database connections automatically

Solution

1. Step 1: Understand serialization concept

   Serialization means converting an object into a format that can be sent or stored, like JSON.

2. Step 2: Relate to Spring Boot usage

   Spring Boot uses serialization to send Java objects as JSON in REST APIs.

3. Final Answer:

   To convert objects into a format like JSON for easy data sharing -> Option C

4. Quick Check:

   Serialization = Convert objects to JSON [OK]

Hint: Serialization means turning objects into JSON or similar formats [OK]

Common Mistakes:

• Confusing serialization with compiling code
• Thinking serialization manages databases
• Mixing serialization with security features

2. Which of the following is the correct way to ensure a Java class is serializable by Spring Boot's default JSON serializer?

easy

A. Use @Entity annotation on the class

B. Make the class implement Serializable interface

C. Mark all fields as transient

D. Add public getters for all fields and a no-argument constructor

Solution

1. Step 1: Identify Spring Boot serialization requirements

   Spring Boot uses Jackson which requires public getters and a no-argument constructor to serialize objects properly.

2. Step 2: Evaluate options

   Implementing Serializable is not required for JSON serialization. Marking fields transient prevents serialization. @Entity is for database mapping, not serialization.

3. Final Answer:

   Add public getters for all fields and a no-argument constructor -> Option D

4. Quick Check:

   Getters + no-arg constructor = JSON serialization [OK]

Hint: Public getters and no-arg constructor enable JSON serialization [OK]

Common Mistakes:

• Thinking Serializable interface is needed for JSON
• Marking fields transient disables serialization
• Confusing @Entity with serialization setup

3. Given the following Spring Boot controller method, what will be the JSON output when accessing /user?

public record User(String name, int age) {}

@GetMapping("/user")
public User getUser() {
    return new User("Alice", 30);
}

medium

A. ["Alice",30]

B. {"name":"Alice","age":30}

C. {"User":{"name":"Alice","age":30}}

D. Error: Records cannot be serialized

Solution

1. Step 1: Understand record serialization in Spring Boot

   Spring Boot with Jackson supports Java records and serializes them as JSON objects with field names.

2. Step 2: Predict output for User record

   The record fields 'name' and 'age' become JSON keys with their values.

3. Final Answer:

   {"name":"Alice","age":30} -> Option B

4. Quick Check:

   Records serialize as JSON objects with fields [OK]

Hint: Java records serialize as JSON objects with field names [OK]

Common Mistakes:

• Expecting JSON array instead of object
• Thinking records cause serialization errors
• Assuming nested JSON with class name

4. Consider this Spring Boot REST controller snippet:

@GetMapping("/product")
public Product getProduct() {
    return new Product("Book", null);
}

public class Product {
    private String name;
    private String description;

    public Product(String name, String description) {
        this.name = name;
        this.description = description;
    }

    public String getName() { return name; }
    // Missing getter for description
}

What issue will occur when this endpoint is called?

medium

A. The JSON will include only the name field; description is missing

B. The JSON will include both name and description fields

C. A runtime error occurs due to missing getter

D. The endpoint will not compile

Solution

1. Step 1: Check serialization requirements

   Jackson serializes fields using public getters. Missing getter means that field is ignored in JSON output.

2. Step 2: Analyze given code

   Product class lacks getter for 'description', so JSON will only show 'name'. No runtime error or compile error occurs.

3. Final Answer:

   The JSON will include only the name field; description is missing -> Option A

4. Quick Check:

   Missing getter = field omitted in JSON [OK]

Hint: Missing getter means field won't appear in JSON output [OK]

Common Mistakes:

• Expecting runtime or compile error
• Assuming all fields serialize without getters
• Thinking null values cause errors

5. You want to customize JSON serialization in Spring Boot to exclude fields with null values from the output. Which approach is correct?

hard

A. Configure Jackson's ObjectMapper to set SerializationInclusion to NON_NULL

B. Mark all nullable fields as transient

C. Remove getters for nullable fields

D. Use @JsonIgnore on all fields

Solution

1. Step 1: Understand how to exclude null fields in JSON

   Jackson allows configuration to exclude null fields by setting SerializationInclusion to NON_NULL.

2. Step 2: Evaluate options

   Marking fields transient or removing getters prevents serialization entirely, not just null exclusion. @JsonIgnore excludes fields always.

3. Final Answer:

   Configure Jackson's ObjectMapper to set SerializationInclusion to NON_NULL -> Option A

4. Quick Check:

   Set SerializationInclusion = NON_NULL to skip nulls [OK]

Hint: Set ObjectMapper to exclude nulls with NON_NULL inclusion [OK]

Common Mistakes:

• Using transient to exclude nulls (excludes always)
• Removing getters breaks serialization
• Using @JsonIgnore excludes fields always