How to Convert Raw JSON to JPA Entities in Spring Boot Using ObjectMapper

Learn how to efficiently convert raw JSON data into JPA entities in Spring Boot REST APIs using Jackson's ObjectMapper, best practices for JSON mapping, and key annotations to handle unknown properties.
---
This video is based on the question https://stackoverflow.com/q/79461230/ asked by the user 'Ziv Aviv' ( https://stackoverflow.com/u/9062178/ ) and on the answer https://stackoverflow.com/a/79461626/ provided by the user 'zn43' ( https://stackoverflow.com/u/5514650/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions.

Visit these links for original content and any more details, such as alternate solutions, latest updates/developments on topic, comments, revision history etc. For example, the original title of the Question was: Spring boot: Creating JPA Entities from raw JSON

Also, Content (except music) licensed under CC BY-SA https://meta.stackexchange.com/help/l...
The original Question post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license, and the original Answer post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license.

If anything seems off to you, please feel free to drop me a comment under this video.
---
Converting Raw JSON to JPA Entities in Spring Boot

When building REST APIs with Spring Boot and Hibernate, a common need is to accept raw JSON from a client and persist it as a JPA entity. The question is: how to convert a JSON map into a strongly-typed entity object efficiently and cleanly?

Common Approach and Its Limitations

A naive approach is to receive the JSON as a Map<String, Object> and manually map each entry to a field in the entity, for example by writing a constructor that extracts and casts each value. However, this quickly becomes unwieldy and error-prone:

Increases boilerplate code

Requires manual type checking

Complicates handling of nested objects or complex types

Recommended Solution: Use Jackson's ObjectMapper

Spring Boot includes Jackson for JSON serialization/deserialization. You can leverage Jackson's ObjectMapper to convert the incoming map directly to your entity class.

Benefits

Clean, concise conversion

Automatic mapping based on field names or @ JsonProperty annotations

Proper handling of nested objects

How to Implement

Add @ JsonIgnoreProperties(ignoreUnknown = true) to your Entity

This annotation prevents errors when the incoming JSON contains fields not present in your entity:

[[See Video to Reveal this Text or Code Snippet]]

Use ObjectMapper.convertValue() in your service layer:

[[See Video to Reveal this Text or Code Snippet]]

Use @ JsonProperty to handle naming mismatches

If your JSON uses different naming conventions (last_name vs. Last_Name), annotate fields in your entity:

[[See Video to Reveal this Text or Code Snippet]]

This ensures the JSON maps correctly to your Java fields.

Additional Best Practices

Avoid using generic Map<String, Object> in controllers: Instead, define DTO classes matching your JSON structure. This allows validation and cleaner code.

Separate endpoints and DTOs for distinct entity types: If your API handles multiple user types or entities, create different DTOs and endpoints to keep concerns clear.

Leverage Bean Validation: Add validation annotations (e.g., @ NotNull, @ Size) on your DTO/entity fields to enforce data integrity automatically.

Summary

Instead of writing manual constructors or mapping code, use Jackson’s ObjectMapper.convertValue() to convert a JSON map into a JPA entity. Annotate your entity with @ JsonIgnoreProperties(ignoreUnknown = true) and use @ JsonProperty for custom mapping. This approach is concise, less error-prone, and fits naturally with Spring Boot’s JSON processing capabilities.