Convert JSON to a Java record (Java 16+)

Records: Immutable Data Carriers

Java 16 introduced record as a first-class language construct for data carrier classes. A record declares its components in a single line, and the compiler synthesizes accessors, equals(), hashCode(), and toString() automatically.

Example JSON

{
  "id": 1,
  "username": "alice",
  "email": "alice@example.com"
}

Generated Java Record

package com.example.model;

public record User(Integer id, String username, String email) {}

Accessors

Records use the field name directly as an accessor (no get prefix):

User user = new User(1, "alice", "alice@example.com");
System.out.println(user.username()); // "alice"

When Records Beat Classes

• DTOs: data transfer between layers
• API responses: immutable, never mutated after deserialization
• Pattern matching (Java 21+): records destructure cleanly in switch expressions
• Map / Set keys: free equals and hashCode make records natural composite keys

When Records Fall Short

• You need mutable fields → records are immutable by design
• You need inheritance → records implicitly extend java.lang.Record and cannot extend other classes
• You need custom setters with validation → records validate via the canonical or compact constructor only
• You target Java 8 / 11 / 14 / 15 → records require Java 16

Custom Constructors

You can validate inputs in the compact constructor without restating the parameter list:

public record User(Integer id, String username, String email) {
    public User {
        Objects.requireNonNull(id, "id");
        if (username.isBlank()) throw new IllegalArgumentException("username");
    }
}

Records and Frameworks

Spring Boot 3+, Jackson 2.12+, and Hibernate 6+ all deserialize directly into records without configuration. Older versions may require additional modules (e.g., jackson-module-parameter-names).