How to Fix Deserialization Issues with New Record Classes in Java?

Why am I unable to deserialize objects when using new Record classes in Java?

// Example Record class in Java
public record Person(String name, int age) { }

Deserialization issues with new Record classes in Java often stem from the unique features of Records, including their immutability and the generated constructors that differ from traditional classes. This guide provides a detailed overview of the causes behind these issues and effective solutions to overcome them.

// Example of using Jackson to deserialize a Record
ObjectMapper mapper = new ObjectMapper();

String json = "{ \"name\": \"John\", \"age\": 30 }";
Person person = mapper.readValue(json, Person.class);

Causes

• Lack of no-argument constructor: Records do not support a default constructor, which can cause issues when deserializing from formats like JSON that expect one.
• Field names case sensitivity: The case of field names in the serialized data must match the record field names exactly, causing mismatches.
• Use of non-serializable fields: If a Record contains a field that is not serializable, an exception will be thrown during deserialization.

Solutions

• Ensure that the serialized data includes all the required fields of the Record.
• Use libraries like Jackson that support deserialization of Records directly by configuring the mapper correctly.
• Make sure that the field names in your serialized data match the exact case of the record field names.

Mistake: Using field names in JSON that do not match the Record's field names.

Solution: Ensure field names in your JSON match the case and naming of the Record's fields.

Mistake: Assuming default constructor is available in a Record.

Solution: Remember that Records in Java do not have a no-argument constructor.