In JPA (Hibernate), when we automatically generate the ID field, it is assumed that the user has no knowledge about this key. So, when obtaining the entity, user would query based on some field other than ID. How do we obtain the entity in that case (since em.find() cannot be used).
I understand we can use a query and filter the results later. But, is there a more direct way (because this is a very common problem as I understand).
I've written a library that helps do precisely this. It allows search by object simply by initializing only the fields you want to filter by: https://github.com/kg6zvp/GenericEntityEJB
Edit: Just realized that @Chinmoy was getting at basically the same thing, but I think I may have done a better job ELI5 :)
If you're using a flavor of Spring Data to help persist / fetch things from whatever kind of Repository you've defined, you can probably have your JPA provider do this for you via some clever tricks with method names in your Repository interface class. Allow me to explain.
(As a disclaimer, I just a few moments ago did/still am figuring this out for myself.)
For example, if I am storing Tokens in my database, I might have an entity class that looks like this:
@Data // << Project Lombok convenience annotation
@Entity
public class Token {
@Id
@Column(name = "TOKEN_ID")
private String tokenId;
@Column(name = "TOKEN")
private String token;
@Column(name = "EXPIRATION")
private String expiration;
@Column(name = "SCOPE")
private String scope;
}
And I probably have a CrudRepository<K,V> interface defined like this, to give me simple CRUD operations on that Repository for free.
@Repository
// CrudRepository<{Entity Type}, {Entity Primary Key Type}>
public interface TokenRepository extends CrudRepository<Token, String> { }
And when I'm looking up one of these tokens, my purpose might be checking the expiration or scope, for example. In either of those cases, I probably don't have the tokenId handy, but rather just the value of a token field itself that I want to look up.
To do that, you can add an additional method to your TokenRepository interface in a clever way to tell your JPA provider that the value you're passing in to the method is not the tokenId, but the value of another field within the Entity class, and it should take that into account when it is generating the actual SQL that it will run against your database.
@Repository
// CrudRepository<{Entity Type}, {Entity Primary Key Type}>
public interface TokenRepository extends CrudRepository<Token, String> {
List<Token> findByToken(String token);
}
I read about this on the Spring Data R2DBC docs page, and it seems to be working so far within a SpringBoot 2.x app storing in an embedded H2 database.
This solution is from Beginning Hibernate book:
Query<User> query = session.createQuery("from User u where u.scn=:scn", User.class);
query.setParameter("scn", scn);
User user = query.uniqueResult();
Write a custom method like this:
public Object findByYourField(Class entityClass, String yourFieldValue)
{
CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<Object> criteriaQuery = criteriaBuilder.createQuery(entityClass);
Root<Object> root = criteriaQuery.from(entityClass);
criteriaQuery.select(root);
ParameterExpression<String> params = criteriaBuilder.parameter(String.class);
criteriaQuery.where(criteriaBuilder.equal(root.get("yourField"), params));
TypedQuery<Object> query = entityManager.createQuery(criteriaQuery);
query.setParameter(params, yourFieldValue);
List<Object> queryResult = query.getResultList();
Object returnObject = null;
if (CollectionUtils.isNotEmpty(queryResult)) {
returnObject = queryResult.get(0);
}
return returnObject;
}
Refer - Spring docs for query methods
We can add methods in Spring Jpa by passing diff params in methods like:
List<Person> findByEmailAddressAndLastname(EmailAddress emailAddress, String lastname);
// Enabling static ORDER BY for a query
List<Person> findByLastnameOrderByFirstnameAsc(String lastname);
All the answers require you to write some sort of SQL/HQL/whatever. Why? You don't have to - just use CriteriaBuilder:
Person.java:
@Entity
class Person {
@Id @GeneratedValue
private int id;
@Column(name = "name")
private String name;
@Column(name = "age")
private int age;
...
}
Dao.java:
public class Dao {
public static Person getPersonByName(String name) {
SessionFactory sessionFactory = new Configuration().configure().buildSessionFactory();
Session session = sessionFactory.openSession();
session.beginTransaction();
CriteriaBuilder cb = session.getCriteriaBuilder();
CriteriaQuery<Person> cr = cb.createQuery(Person.class);
Root<Person> root = cr.from(Person.class);
cr.select(root).where(cb.equal(root.get("name"), name)); //here you pass a class field, not a table column (in this example they are called the same)
Query<Person> query = session.createQuery(cr);
query.setMaxResults(1);
List<Person> result = query.getResultList();
session.close();
return result.get(0);
}
}
example of use:
public static void main(String[] args) {
Person person = Dao.getPersonByName("John");
System.out.println(person.getAge()); //John's age
}
Using CrudRepository and JPA query works for me:
import org.springframework.data.jpa.repository.Query;
import org.springframework.data.repository.CrudRepository;
import org.springframework.data.repository.query.Param;
public interface TokenCrudRepository extends CrudRepository<Token, Integer> {
/**
* Finds a token by using the user as a search criteria.
* @param user
* @return A token element matching with the given user.
*/
@Query("SELECT t FROM Token t WHERE LOWER(t.user) = LOWER(:user)")
public Token find(@Param("user") String user);
}
and you invoke the find custom method like this:
public void destroyCurrentToken(String user){
AbstractApplicationContext context = getContext();
repository = context.getBean(TokenCrudRepository.class);
Token token = ((TokenCrudRepository) repository).find(user);
int idToken = token.getId();
repository.delete(idToken);
context.close();
}
Best practice is using @NaturalId annotation. It can be used as the business key for some cases it is too complicated, so some fields are using as the identifier in the real world. For example, I have user class with user id as primary key, and email is also unique field. So we can use email as our natural id
@Entity
@Table(name="user")
public class User {
@Id
@Column(name="id")
private int id;
@NaturalId
@Column(name="email")
private String email;
@Column(name="name")
private String name;
}
To get our record, just simply use 'session.byNaturalId()'
Session session = sessionFactory.getCurrentSession();
User user = session.byNaturalId(User.class)
.using("email","[email protected]")
.load()
if you have repository for entity Foo and need to select all entries with exact string value boo (also works for other primitive types or entity types). Put this into your repository interface:
List<Foo> findByBoo(String boo);
if you need to order results:
List<Foo> findByBooOrderById(String boo);
See more at reference.
Basically, you should add a specific unique field. I usually use xxxUri fields.
class User {
@Id
// automatically generated
private Long id;
// globally unique id
@Column(name = "SCN", nullable = false, unique = true)
private String scn;
}
And you business method will do like this.
public User findUserByScn(@NotNull final String scn) {
CriteriaBuilder builder = manager.getCriteriaBuilder();
CriteriaQuery<User> criteria = builder.createQuery(User.class);
Root<User> from = criteria.from(User.class);
criteria.select(from);
criteria.where(builder.equal(from.get(User_.scn), scn));
TypedQuery<User> typed = manager.createQuery(criteria);
try {
return typed.getSingleResult();
} catch (final NoResultException nre) {
return null;
}
}
In my Spring Boot app I resolved a similar type of issue like this:
@Autowired
private EntityManager entityManager;
public User findByEmail(String email) {
User user = null;
Query query = entityManager.createQuery("SELECT u FROM User u WHERE u.email=:email");
query.setParameter("email", email);
try {
user = (User) query.getSingleResult();
} catch (Exception e) {
// Handle exception
}
return user;
}
No, you don't need to make criteria query it would be boilerplate code you just do simple thing if you working in Spring-boot: in your repo declare a method name with findBy[exact field name]. Example- if your model or document consist a string field myField and you want to find by it then your method name will be:
findBymyField(String myField);
It is not a "problem" as you stated it.
Hibernate has the built-in find(), but you have to build your own query in order to get a particular object. I recommend using Hibernate's Criteria :
Criteria criteria = session.createCriteria(YourClass.class);
YourObject yourObject = criteria.add(Restrictions.eq("yourField", yourFieldValue))
.uniqueResult();
This will create a criteria on your current class, adding the restriction that the column "yourField" is equal to the value yourFieldValue. uniqueResult() tells it to bring a unique result. If more objects match, you should retrive a list.
List<YourObject> list = criteria.add(Restrictions.eq("yourField", yourFieldValue)).list();
If you have any further questions, please feel free to ask. Hope this helps.
Have a look at:
JPA query language: The Java Persistence Query Language JPA Criteria API: Using the Criteria API to Create QueriesI solved a similar problem, where I wanted to find a book by its isbnCode not by your id(primary key).
@Entity
public class Book implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
private Integer id;
private String isbnCode;
...
In the repository the method was created like @kamalveer singh mentioned. Note that the method name is findBy+fieldName (in my case: findByisbnCode):
@Repository
public interface BookRepository extends JpaRepository<Book, Integer> {
Book findByisbnCode(String isbnCode);
}
Then, implemented the method in the service:
@Service
public class BookService {
@Autowired
private BookRepository repo;
public Book findByIsbnCode(String isbnCode) {
Book obj = repo.findByisbnCode(isbnCode);
return obj;
}
}
Source: Stackoverflow.com