I would like to know the difference between JPA and Hibernate. I have read the very interesting question posted by #Anthony with interest but still I do not understand the full picture.
I have implemented my application in Spring MVC and Hibernate (see below). My DAOs are an implementation of services which have been built using HQL queries.
#Service("messagesService")
public class MessagesService
{
private MessagesDAO messagesDAO;
#Autowired
public void setMessagesDAO(MessagesDAO messagesDAO)
{
this.messagesDAO = messagesDAO;
}
public List<Message> getAllMessages()
{
return messagesDAO.getAllMessages();
}
...
--------
import org.hibernate.Criteria;
import org.hibernate.Query;
import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.criterion.Restrictions;
import org.springframework.stereotype.Component;
import org.springframework.stereotype.Repository;
import org.springframework.transaction.annotation.Transactional;
#Repository
#Transactional
#Component("messagesDAO")
public class MessagesDAO
{
#Autowired
private SessionFactory sessionFactory;
public Session session()
{
return sessionFactory.getCurrentSession();
}
#SuppressWarnings("unchecked")
public List<Message> getAllMessages()
{
Criteria crit = session().createCriteria(Message.class);
crit.createAlias("usernameSender", "u").add(Restrictions.eq("u.enabled",true));
return crit.list();
}
...
I really like the statement "JPA is the dance, Hibernate is the dancer." but in my specific case I do not fully get why my example is not JPA. MessageService is the dance and MessagesDAO the dancer(Implementation).
As #Kevin states:
Think of JPA as the guidelines that must be followed or an interface,
while Hibernate's JPA implementation is code that meets the API as
defined by the JPA specification and provides the under the hood
functionality.
I know I have not defined my service as an interface but this still lets me think that my code complies with the JPA specification requirements.
Now a concern arises
What is the difference between my example and the following from the pet clinic example
package org.springframework.samples.petclinic.repository;
import java.util.List;
import org.springframework.dao.DataAccessException;
import org.springframework.samples.petclinic.model.BaseEntity;
import org.springframework.samples.petclinic.model.Pet;
import org.springframework.samples.petclinic.model.PetType;
public interface PetRepository {
List<PetType> findPetTypes() throws DataAccessException;
-------
#Repository
public class JpaPetRepositoryImpl implements PetRepository {
#PersistenceContext
private EntityManager em;
#Override
#SuppressWarnings("unchecked")
public List<PetType> findPetTypes() {
return this.em.createQuery("SELECT ptype FROM PetType ptype ORDER BY ptype.name").getResultList();
}
The reason why I am asking all this questions is because I am using MySql in my application and I am thinking to change it in the future.
Therefore, I am trying to build my implementation layer to avoid any problem later on.
I was looking at nosql options and I discovered spring data jpa for the integration layer. I then started to learn a bit more about JPA and DAOs and the questions above suddenly arose
If I implement spring data jpa, can I use MySql for the moment and change it later to another database (Cassandra, MongoDb)?
What is the difference between Spring data JPA and Spring Data Mongodb
(The first is the specification and the second one the implementation?)
Thank you for your help
To keep the terminology: In your DAO you can't change your dancing partner.
Why? Because you are referencing Hibernate explicitly. If you would like to change the dancer some day you would have to change your whole implementation. Thats why you usually use only classes of the dance - JPA or Spring Data in your case.
Related
I've been searching for the simplest and best way to validate my entities before they are created/updated and after much googling, I couldn't find a clean/modern one.
Ideally, I would have loved to be able to use #Valid as follows:
import javax.validation.Valid;
import lombok.extern.slf4j.Slf4j;
import org.springframework.data.rest.core.annotation.HandleBeforeCreate;
import org.springframework.data.rest.core.annotation.HandleBeforeSave;
import org.springframework.data.rest.core.annotation.RepositoryEventHandler;
import org.springframework.stereotype.Component;
import org.springframework.validation.annotation.Validated;
#Slf4j
#Validated
#Component
#RepositoryEventHandler
public class CustomerEventHandler {
// Triggered for POST
#HandleBeforeCreate
public void onBeforeCreate(#Valid Customer entity) {
log.info("Saving new entity {}", entity);
}
// Triggered for PUT / PATCH
#HandleBeforeSave
public void onBeforeSave(#Valid Customer entity) {
log.info("Saving new entity {}", entity);
}
}
The Customer entity being:
import javax.validation.constraints.NotBlank;
#Getter
#Setter
#ToString
#AllArgsConstructor
#NoArgsConstructor
#Entity
#Table(name = "customer")
public class Customer {
#NotBlank
private String firstname;
}
But it doesn't seem to work.
What's the modern, easy way to validate entities in Spring Data REST?
Note: I'm using Spring Boot
I checked your pom.xml in linked GitHub project. You have just a dependency to validation annotations, but the proper way with Spring Boot is to use the spring-boot-starter-validation Dependency. The Spring Boot Starter Dependencies add the "magic" to your project, which triggers automatically the validation based on your annotations.
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-validation</artifactId>
</dependency>
I my German blog I have written an article about this topic:
https://agile-coding.blogspot.com/2020/11/validation-with-spring.html
I want to suggest a few best practise that every developer, who starting as junior/beginner should be know. Don't put any Validation annotation in Entities, using them in DTO/Resource classes. and the best way to practise validation is that you can handler MethodArgumentNotValidation exception in your own Spring Boot of Exception Handler class annotated that class #RestControllerAdvice and create your own #interface annotation instead of using more validation annotation.
I am working on Spring Boot application.
We have Service Layer,Rest Controller and Dao as repository.
I have 20 to 30 tables in my database and I dont want to create repository for each entity and extends that to CrudRepository.
ex : User is an Entity, to perform persistance operations on User, I have to create UserRepository which extends CrudRepository.
Same with Department, Company etc...
What i want to do is, I will write a BaseRepository which gonna extend CrudRepository, base repository should accept all entities and do persistance operations.
Is there a way to that ??
Don't extend CrudRepository it's functionality is all tied to the generic type, it'd be hacky to extend it for a generic implementation. You probably just want something simple which uses the JPA entity manager directly:
import javax.persistence.EntityManager;
import org.springframework.beans.factory.annotation.Autowired;
public class GenericRepository {
#Autowired
private EntityManager entityManager;
public <T, ID> T findById(Class<T> type, ID id) {
return entityManager.find(type, id);
}
}
i have a Spring Boot project which has some external packages i need to import as Beans in the main application.
So i have my main application in com.package.app package and some classes (among which some repositories) in com.package.commons package.
In order to take these beans i have my main class annotated as follows:
#SpringBootApplication
#ComponentScan({ "com.package.commons" ,"com.package.app"})
#EnableScheduling
#EnableAsync
public class EmanagerApplication extends SpringBootServletInitializer{
public static void main(String[] args) {
SpringApplication.run(EmanagerApplication.class, args);
}
}
But when i launch the application it may occur (not always but very ofter) that the start up fails with these kind of error:
Description:
Field repository in com.package.commons.service.BrandService required a bean of type 'com.package.commons.persistence.repository.BrandRepository' that could not be found.
Action:
Consider defining a bean of type 'com.package.commons.persistence.repository.BrandRepository' in your configuration.
My BrandRepository is annotated with #Repository and the service class with #Service
The really strange thing is that if i keep launching the app at the end it stars... but there is no reason for it...
If you're using JPA, you'll also need the #EnableJpaRepositories annotation.
Also consider to use #EnableTransactionManagement to enable declarative transaction handling.
E.g. use something like the following in the same package or a parent package where you have your JPA entities and JPA repositories (untested):
import java.util.HashMap;
import java.util.Map;
import javax.sql.DataSource;
import org.springframework.beans.factory.ObjectProvider;
import org.springframework.boot.autoconfigure.orm.jpa.JpaBaseConfiguration;
import org.springframework.boot.autoconfigure.orm.jpa.JpaProperties;
import org.springframework.boot.autoconfigure.transaction.TransactionManagerCustomizers;
import org.springframework.orm.jpa.vendor.AbstractJpaVendorAdapter;
import org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter;
import org.springframework.transaction.jta.JtaTransactionManager;
#Configuration
#EntityScan
#EnableJpaRepositories
#EnableTransactionManagement
public class HibernateConfig extends JpaBaseConfiguration {
public HibernateConfig(DataSource dataSource, JpaProperties properties, ObjectProvider<JtaTransactionManager> jtaTransactionManager,
ObjectProvider<TransactionManagerCustomizers> transactionManagerCustomizers) {
super(dataSource, properties, jtaTransactionManager, transactionManagerCustomizers);
}
#Override
protected AbstractJpaVendorAdapter createJpaVendorAdapter() {
return new HibernateJpaVendorAdapter();
}
#Override
protected Map<String, Object> getVendorProperties() {
return new HashMap<>();
}
}
And don't forget to annotate your #Service classes also with #Transactional.
If you confirm that the Application which with the startup method of this application is good, and confirm the #ComponentScan is good also. And the configuration file yaml or properties of JPA also good.
How about trying extends JPA Repository like this:
public class xxxResponsitory extends JpaRepository<T, E>{
...
}
Cause JpaRepository has already annotated with #Repository annotation, T means the type of Primary Key, I always use Integer or Long, autoboxing type. E means the main type of this repository.
Make an example:
Now we have an Entity type named User, the Primary key type of User is Long, I would write the repository like this:
public class UserRepository extends JpaRepository<Long, User>{
...
}
Don't need annotated anything, then, In the service class, #Autowried UserRepository, everything is good to run. But make sure the things that I talk at the start of my answer.
Hope this can help you.
I am trying to use Spring data and repositories in a Spring Boot application, but I have an error when compiling the project.
Here is my Entity :
package fr.investstore.model;
import javax.persistence.Id;
...
#Entity
public class CrowdOperation {
#Id
#GeneratedValue(strategy=GenerationType.IDENTITY)
public Long id;
#Enumerated(EnumType.STRING)
public RepaymentType repaymentType;
...
}
And the corresponding Repository:
package fr.investstore.repositories;
import org.springframework.data.repository.CrudRepository;
import org.springframework.data.repository.PagingAndSortingRepository;
import org.springframework.stereotype.Repository;
import fr.investstore.model.CrowdOperation;
public interface CrowdOperationRepository extends CrudRepository<CrowdOperation, Long> {
}
I use it in a WS controller, generating a repository through the Autowired annotation:
package fr.investstore.ws;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.boot.autoconfigure.EnableAutoConfiguration;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.ResponseBody;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.bind.annotation.RequestMapping;
...
#Controller
#EnableAutoConfiguration
public class SampleController {
#Autowired
private CrowdOperationRepository crowdOperationRepository;
#RequestMapping(path = "/", method = RequestMethod.GET)
#ResponseBody
public String getOperations(#RequestParam(required=true, defaultValue="Stranger") String name) {
crowdOperationRepository.save(new CrowdOperation());
return "Hello " + name;
}
}
And the code of the application:
package fr.investstore;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import fr.investstore.ws.SampleController;
#SpringBootApplication
public class InvestStoreApplication {
public static void main(String[] args) {
SpringApplication.run(SampleController.class, args);
}
}
But when compiling the project I get:
APPLICATION FAILED TO START
Description: Field crowdOperationRepository in
fr.investstore.ws.SampleController required a bean of type
'fr.investstore.repositories.CrowdOperationRepository' that could not
be found.
Action: Consider defining a bean of type
'fr.investstore.repositories.CrowdOperationRepository' in your
configuration.
Woudn't Spring automatically generate a bean for the repository through the interface?
How can I resolve this?
EDIT: I also tried to put the Repository annotation (from org.springframework.stereotype.Repository) onto CrowdOperationRepository, but I got the same error
While creating a spring-boot application, we need to keep some point in our mind like
Always keep main class (class with `#SpringBootApplication annotation) on the top level package and other classes should lie under sub-packages.
Always mark your bean classes with proper annotation e.g. all repositories should be marked by #Repository annotation, all service implementation classes should be marked with #Service, other component classes should be marked by #Component, class which defines our beans should be marked as #Configuration
Enable the feature which you are using e.g. #EnableJpaRepositories, #EnableTransactionManagement, #EnableJpaAuditing, these annotations also provides functionality which let us define which package spring needs to scan.
So in your case, you need to mark InvestStoreApplication class with #EnableJpaRepositories annotation and CrowdOperationRepository with #Repository.
you have to tell your spring boot application to load JPA repositories.
copy this one to your application class
it will auto-scan your JPA repository and load it in your spring container even if you do not define your interface with #Repository it will wire that bean in your dependent class.
#EnableJpaRepositories(basePackages = { "fr.investstore.repositories" })
Thank to #JBNizet for his comment, that made it working.
I create this answer since he did not:
Replace SpringApplication.run(SampleController.class, args); with SpringApplication.run(InvestStoreApplication.class, args);. And remove the useless #EnableAutoConfiguration on your controller.
Annotating your entity class as shown as spring hint below to allow spring get a valid repository bean
Spring Data JPA - Could not safely identify store assignment for repository candidate interface com.xxxxx.xxxxRepository.
If you want this repository to be a JPA repository, consider annotating your entities with one of these annotations: javax.persistence.Entity, javax.persistence.MappedSuperclass (preferred),
or consider extending one of the following types with your repository: org.springframework.data.jpa.repository.JpaRepository.
2022-05-06 12:32:12.623 [ restartedMain] INFO [.RepositoryConfigurationDelegate:201 ] - Finished Spring Data repository scanning in 3 ms. Found 0 JPA repository interfaces.
everything is in the title...
I have a repository extending JpaRepository<User, Long> with additional customized methods, following the pattern described in http://docs.spring.io/spring-data/jpa/docs/1.4.3.RELEASE/reference/html/jpa.repositories.html
The Spring documentation says: CRUD methods on repository instances are transactional by default. For reading operations the transaction configuration readOnly flag is set to true, all others are configured with a plain #Transactional.
Does this applies to the custom-methods? Are there transactional?
Advanced: Where could I find the answer by myself? (i.e. which class of Spring framework is responsible for generating the dynamic-proxy and decorating the calls?)
Thank you in advance!
1) Same as Ralphs answer. They have to be annotated with #Transaction or handled by you explicitly
2) There is no "responsible" class for extending/decorating the Spring repository interfaces. YOU are the one responsible for extending them.
I'm going to add an example of a custom method that executes a custom query.
PersonRepository
public interface PostRepository extends CrudRepository<Post, Long>{
List<Post> getPostsOlderThanDate(Date date);
}
PersonRepositoryImpl
import org.springframework.data.jpa.repository.support.SimpleJpaRepository;
import javax.persistence.EntityManager;
import javax.persistence.TypedQuery;
import java.util.Date;
import java.util.List;
public class PostRepositoryImpl extends SimpleJpaRepository<Post, Long> implements PostRepository {
private EntityManager entityManager;
public PostRepositoryImpl(EntityManager em) {
super(Post.class, em);
this.entityManager = em;
}
#Transactional
#Override
public List<Post> getPostsOlderThanDate(Date date) {
String query = ""; //create query
TypedQuery<Post> typedQuery = entityManager.createQuery(query, Post.class);
return typedQuery.getResultList();
}
}
You then create your EntityManagerFactory bean (which you build with a Hibernate persistance provider, or however you set it up, it's up to you), and inject it into the constructor of the PersonRepositoryImpl bean.
<bean id="personRepository" class="com.package.PersonRepositoryImpl">
<constructor-arg ref="entityManagerFactory" />
</bean>
These would be the steps to provide some custom queries into your database.
NOTE: if you need help setting up your EntityManager just leave a comment addressed to me and I'll give you the spring config and dependencies.