Spring Boot Auditing - map current user to @CreatedBy, @LastModifiedBy

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So I have an audit class that is using @MappedSuperClass, it updates the values createdBy and updatedBy but it does not add a foreign key on the User entity, so there's no database validation

Here's the Audit Class


@Getter
@Setter
@MappedSuperclass
@JsonIgnoreProperties(
        value = {"createdBy", "updatedBy"},
        allowGetters = true
)
public abstract class UserDateAudit extends DateAudit {

    @CreatedBy
    @Column(name = "created_by", nullable = false, updatable = false)
    public Long createdBy;

    @LastModifiedBy
    @Column(name = "updated_by", nullable = false)
    public Long updatedBy;
}

I searched on how to do this using @Inheritance but did not completely get it. so How do I achieve the connection between this class and User entity?

Edit 1 Here's the auditing configuration I implement.

@Configuration
@EnableJpaAuditing
public class AuditingConfig {

    @Bean
    public AuditorAware<Long> auditorProvider() {
        return new SpringSecurityAuditAwareImpl();
    }

}

class SpringSecurityAuditAwareImpl implements AuditorAware<Long> {

    @Override
    public Optional<Long> getCurrentAuditor() {
        Authentication authentication = SecurityContextHolder.getContext().getAuthentication();

        if (authentication == null ||
                !authentication.isAuthenticated() ||
                authentication instanceof AnonymousAuthenticationToken) {
            return Optional.empty();
        }

        UserPrincipal userPrincipal = (UserPrincipal) authentication.getPrincipal();

        return Optional.ofNullable(userPrincipal.getId());
    }
}

Edit 2 to set clear what exactly I mean so how to use something other than @MappedSuperclass to be able achieve this "I want to be able to map the user reference in all tables that inherit the UserDateAudit so that it is a foreign key for all these table (which would add a validation that the user id actually exist) not just a regular column".

spring-boot jpa hibernate spring java
1 Answers

You need to implement the org.springframework.data.domain.AuditorAware interface and add the logic to retrieve currently logged in user in the getCurrentAuditor method.

@Component
@RequiredArgsConstructor
public class AuditorResolver implements AuditorAware<YOUR_TYPE> {
 
    @Override
    public Optional<YOUR_TYPE> getCurrentAuditor() {

        //code to retrieve the currently logged in user and return the id/user object 
    }
}

See this for an example: https://github.com/gtiwari333/spring-boot-blog-app/blob/master/src/main/java/gt/app/config/AuditorResolver.java

The documentation also describes it beautifully: https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#auditing.basics

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