Message serialization in Spring Boot - Step-by-Step Execution

Concept Flow - Message serialization

Create Java Object

↓

Serialize Object to JSON

↓

Send JSON over Network

↓

Receive JSON

↓

Deserialize JSON to Java Object

↓

Use Java Object in Application

This flow shows how a Java object is converted to JSON for sending, then back to a Java object when received.

Execution Sample

Spring Boot

import com.fasterxml.jackson.databind.ObjectMapper;

public class User {
  private String name;
  private int age;

  public User() {}

  public User(String name, int age) {
    this.name = name;
    this.age = age;
  }

  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;
  }
}

User user = new User("Alice", 30);
String json = new ObjectMapper().writeValueAsString(user);

This code creates a User object and converts it to a JSON string using Jackson's ObjectMapper.

Execution Table

Step	Action	Input	Output	Notes
1	Create User object	name="Alice", age=30	User{name='Alice', age=30}	Java object created in memory
2	Serialize to JSON	User object	{"name":"Alice","age":30}	ObjectMapper converts object to JSON string
3	Send JSON	{"name":"Alice","age":30}	JSON string sent over network	JSON is text format suitable for transfer
4	Receive JSON	JSON string	JSON string received	Data arrives as JSON text
5	Deserialize JSON	{"name":"Alice","age":30}	User{name='Alice', age=30}	ObjectMapper converts JSON back to Java object
6	Use Java object	User object	Access user.name and user.age	Application uses deserialized object
7	End	N/A	N/A	Process complete

💡 Serialization and deserialization complete, object ready for use

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 5	Final
user	null	User{name='Alice', age=30}	User{name='Alice', age=30}	User{name='Alice', age=30}	User{name='Alice', age=30}
json	null	null	{"name":"Alice","age":30}	{"name":"Alice","age":30}	{"name":"Alice","age":30}

Key Moments - 2 Insights

Why do we convert a Java object to JSON before sending it?

Because JSON is a text format that can be sent over networks easily, unlike Java objects which only exist in memory (see execution_table step 2 and 3).

What happens if the JSON structure does not match the Java class during deserialization?

Deserialization will fail or produce incorrect objects because ObjectMapper expects JSON fields to match Java class fields (refer to execution_table step 5).

Visual Quiz - 3 Questions

Look at the execution table, what is the value of 'json' after step 2?

A User{name='Alice', age=30}

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

C null

D JSON object

💡 Hint

Check the 'Output' column at step 2 in the execution_table.

At which step does the Java object get recreated from JSON?

A Step 5

B Step 3

C Step 2

D Step 4

💡 Hint

Look for 'Deserialize JSON' action in the execution_table.

If the 'user' object had an extra field not in JSON, what would happen during deserialization?

A Deserialization ignores the extra field and succeeds

B Deserialization fails with error

C Extra field gets default value

D JSON adds the extra field automatically

💡 Hint

Think about how ObjectMapper handles missing JSON fields during deserialization (see key_moments).

Concept Snapshot

Message serialization in Spring Boot:
- Convert Java object to JSON string using ObjectMapper
- Send JSON over network as text
- Receive JSON and convert back to Java object
- JSON must match Java class structure
- Enables data exchange between systems

Full Transcript

Message serialization means turning a Java object into a JSON string so it can be sent over a network. In Spring Boot, we use ObjectMapper to do this. First, we create a Java object like a User with name and age. Then ObjectMapper converts it to JSON text. This JSON is sent over the network. When received, ObjectMapper reads the JSON and creates a new Java object with the same data. This process allows different systems to share data easily. The JSON structure must match the Java class fields to work correctly.

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