I am working with Hibernate 4+ Spring MVC + Spring Data with JPA annotations:
#Entity
public class ClassOne implements Serializable{
......
#OneToMany(mappedBy = "mapper", fetch=FetchType.LAZY)
private Set<ClassTwo> element = new HashSet<ClassTwo>(0);
//more fields
//getters and setters
//equals and hashcode
}
#Entity
public class ClassTwo implements Serializable{
......
#ManyToOne(fetch=FetchType.LAZY)
#JoinColumn(name = "CEN_CEN_CODIGO", nullable = false)
private ClassOne classOne;
//more fields
//getters and setters
//equals and hashcode
}
public interface ClassOneRepository extends CrudRepository<ClassOne, Long> {
#Override
#Query("select c from ClassOne c")
public List<ClassOne> findAll();
}
#Service
public class ClassOneService {
#Autowired
private ClassOneRepository classOneRepository;
#Transactional(readOnly=true)
public List<ClassOne> findAll() {
return classOneRepository.findAll();
}
}
And finally I call thie service from my #Controller
#Autowired
ClassOneService classOneService;
I expect results ONLY from ClassOne but retrieving the JOIN values with ClassTwo and all the database tree associate. Is it possible to get only values for ONE table using this schema? Is it a cache problem or Fetching not LAZY?
EDIT: I added the relatioship between two classes
Thank you
You must have the following anotation above your Set<ClassTwo> or its getter:
#OneToMany(fetch = FetchType.LAZY, ...)
See http://docs.oracle.com/javaee/7/api/javax/persistence/OneToMany.html#fetch()
It seems to be that simple "SELECT *" JPA query returns all eagerly fetched fields for the entity.
Please refer to: JPA - Force Lazy loading for only a given query
and http://forcedotcom.github.io/java-sdk/jpa-queries.
So my solution would be to use SessionFactory to get current session and then use "classic" method
return getCurrentSession().createCriteria(persistentClass).list();
Another possible way is to create let's say a POJO object without Set which uses the same table as ClassOne.
I've just added #Lazy for each #ManyToOne and #OneToMany relationship. It seems that Spring Data needs Spring annotations but I supposed that just was necessary to add fetch = FetchType.LAZY. No more Eager behaviours ;).
Thanks for your responses
Related
I am trying to implement the OneToOne association in JPA and trying to join two tables using spring boot and spring data JPA. I created one spring boot microservice and implemented the one to one association in my model. But when I am running code I am getting the following error ,
Caused by: org.hibernate.AnnotationException: Illegal attempt to map a non collection as a #OneToMany, #ManyToMany or #CollectionOfElement
Here My First model class Users.java is like following,
#Entity
#Table(name = "users")
public class Users implements Serializable {
private static final long serialVersionUID = 9178661439383356177L;
#Id
#Column(name="user_id")
public Integer userId;
#Column(name="username")
public String username;
#Column(name="password")
public String password;
}
And I am testing association by controller using following code,
#GetMapping("/load")
public Users load() {
return (Users) userObj.findAll();
}
Can anyone help to resolve this association issue please ?
This is wrong.
#OneToOne(mappedBy="nuserId")
public Set<UserRoleMapping> roleUserRoleMappingMappingJoin;
}
OneToOne means only one object..right?
See this for mappings understandings.
https://docs.jboss.org/hibernate/orm/3.6/reference/en-US/html/collections.html#collections-persistent
Annotation #OneToOne defines a single-valued association to another entity, and in your case you associate a user to a Set of UserRoleMapping instead of associating it with a single object of that class. Use #ManyToOne annotation
Actually the exception refers to an invalid #OneToMany, #ManyToMany or #CollectionOfElement mapping
and this can only be
#OneToMany()
#JoinColumn(name="nuser_id" , referencedColumnName="nuserId")
public Users nuserId;
If the #OneToMany relation is valid change this at first to
#OneToMany()
#JoinColumn(name="nuser_id" , referencedColumnName="nuserId")
public List<Users> users;
If the #OneToMany relation is NOT valid change this to
#OneToOne()
#JoinColumn(name="nuser_id" , referencedColumnName="nuserId")
public Users users;
I have a database service using Spring Boot 1.5.1 and Spring Data Rest. I am storing my entities in a MySQL database, and accessing them over REST using Spring's PagingAndSortingRepository. I found this which states that sorting by nested parameters is supported, but I cannot find a way to sort by nested fields.
I have these classes:
#Entity(name = "Person")
#Table(name = "PERSON")
public class Person {
#ManyToOne
protected Address address;
#ManyToOne(targetEntity = Name.class, cascade = {
CascadeType.ALL
})
#JoinColumn(name = "NAME_PERSON_ID")
protected Name name;
#Id
protected Long id;
// Setter, getters, etc.
}
#Entity(name = "Name")
#Table(name = "NAME")
public class Name{
protected String firstName;
protected String lastName;
#Id
protected Long id;
// Setter, getters, etc.
}
For example, when using the method:
Page<Person> findByAddress_Id(#Param("id") String id, Pageable pageable);
And calling the URI http://localhost:8080/people/search/findByAddress_Id?id=1&sort=name_lastName,desc, the sort parameter is completely ignored by Spring.
The parameters sort=name.lastName and sort=nameLastName did not work either.
Am I forming the Rest request wrong, or missing some configuration?
Thank you!
The workaround I found is to create an extra read-only property for sorting purposes only. Building on the example above:
#Entity(name = "Person")
#Table(name = "PERSON")
public class Person {
// read only, for sorting purposes only
// #JsonIgnore // we can hide it from the clients, if needed
#RestResource(exported=false) // read only so we can map 2 fields to the same database column
#ManyToOne
#JoinColumn(name = "address_id", insertable = false, updatable = false)
private Address address;
// We still want the linkable association created to work as before so we manually override the relation and path
#RestResource(exported=true, rel="address", path="address")
#ManyToOne
private Address addressLink;
...
}
The drawback for the proposed workaround is that we now have to explicitly duplicate all the properties for which we want to support nested sorting.
LATER EDIT: another drawback is that we cannot hide the embedded property from the clients. In my original answer, I was suggesting we can add #JsonIgnore, but apparently that breaks the sort.
I debugged through that and it looks like the issue that Alan mentioned.
I found workaround that could help:
Create own controller, inject your repo and optionally projection factory (if you need projections). Implement get method to delegate call to your repository
#RestController
#RequestMapping("/people")
public class PeopleController {
#Autowired
PersonRepository repository;
//#Autowired
//PagedResourcesAssembler<MyDTO> resourceAssembler;
#GetMapping("/by-address/{addressId}")
public Page<Person> getByAddress(#PathVariable("addressId") Long addressId, Pageable page) {
// spring doesn't spoil your sort here ...
Page<Person> page = repository.findByAddress_Id(addressId, page)
// optionally, apply projection
// to return DTO/specifically loaded Entity objects ...
// return type would be then PagedResources<Resource<MyDTO>>
// return resourceAssembler.toResource(page.map(...))
return page;
}
}
This works for me with 2.6.8.RELEASE; the issue seems to be in all versions.
From Spring Data REST documentation:
Sorting by linkable associations (that is, links to top-level resources) is not supported.
https://docs.spring.io/spring-data/rest/docs/current/reference/html/#paging-and-sorting.sorting
An alternative that I found was use #ResResource(exported=false).
This is not valid (expecially for legacy Spring Data REST projects) because avoid that the resource/entity will be loaded HTTP links:
JacksonBinder
BeanDeserializerBuilder updateBuilder throws
com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of ' com...' no String-argument constructor/factory method to deserialize from String value
I tried activate sort by linkable associations with help of annotations but without success because we need always need override the mappPropertyPath method of JacksonMappingAwareSortTranslator.SortTranslator detect the annotation:
if (associations.isLinkableAssociation(persistentProperty)) {
if(!persistentProperty.isAnnotationPresent(SortByLinkableAssociation.class)) {
return Collections.emptyList();
}
}
Annotation
#Retention(RetentionPolicy.RUNTIME)
#Target(ElementType.FIELD)
public #interface SortByLinkableAssociation {
}
At project mark association as #SortByLinkableAssociation:
#ManyToOne
#SortByLinkableAssociation
private Name name;
Really I didn't find a clear and success solution to this issue but decide to expose it to let think about it or even Spring team take in consideration to include at nexts releases.
Please see https://stackoverflow.com/a/66135148/6673169 for possible workaround/hack, when we wanted sorting by linked entity.
I'm using Spring JPA in my DAO layer. I have an entity Projet having inside an entity property Client:
Project.java
#Entity
public class Project {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private int projetId;
private String libelle;
#OneToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
#JoinColumn(name="client_id")
private Client client;
// ... constructors, getters & setters
}
Client.java
#Entity
public class Client {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private int clientId;
private String denomination;
// ... constructors, getters & setters
}
in my DAO interface I have the following specifications:
ProjetDao.java
#Repository
#Transactional
public interface ProjetDao extends CrudRepository<Projet, Integer> {
#Transactional
public Projet findByLibelle(String libelle);
#Transactional
public Projet findByProjetId(int projetId);
}
My question is: How can I specify in my DAO interface a method that will return all clients distinct in List<Client>?
From the documentation and JIRA:
List<Project> findAllDistinctBy();
The query builder mechanism built into Spring Data repository infrastructure is useful for building constraining queries over entities of the repository. The mechanism strips the prefixes find…By, read…By, query…By, count…By, and get…By from the method and starts parsing the rest of it. The introducing clause can contain further expressions such as a Distinct to set a distinct flag on the query to be created. However, the first By acts as delimiter to indicate the start of the actual criteria. At a very basic level you can define conditions on entity properties and concatenate them with And and Or.
You are dealing with a one-to-one relationship, in this case I guess the list that you need is not really related to specific project, you just want a distinct list of clients.
You will need to create another repository (ClientRepository) for the Client entity and add a findAllDistinct method in this repository.
is it possible to have a projection with nested collection with Spring JPA?
I have the following 2 simple entity (to explain the problem)
#Entity
#Table(name = "person")
public class Person implements Serializable {
private Integer id;
private String name;
#OneToMany
private List<Address> addressList = new ArrayList<>();
}
#Entity
#Table(name = "address")
public class Address implements Serializable {
private Integer id;
private String city;
private String street;
}
Is it possible to have a projection of Person with following attributes filled in ? {person.name, address.city}
I might be wrong in semantics of word Projection. but the problem is what i need to achieve. Maybe it is not possible with Projection, but is there another way to achieve the end goal? Named Entity graph perhaps ?
P.S. please suggest a solution for Spring JPA not Spring Jpa REST
thanks in advance
You're right, Entity Graphs serve this exact purpose - control field loading.
Create entity graphs dynamically from the code or annotate target entities with Named Entity Graphs and then just use their name.
Here is how to modify your Person class to use Named Entity Graphs:
#Entity
#Table(name = "person")
#NamedEntityGraph(name = "persion.name.with.city",
attributeNodes = #NamedAttributeNode(value = "addressList", subgraph = "addresses.city"),
subgraphs = #NamedSubgraph(name = "addresses.city", attributeNodes = #NamedAttributeNode("city")))
public class Person implements Serializable {
private Integer id;
private String name;
#OneToMany
private List<Address> addressList;
}
And then when loading your person:
EntityGraph graph = em.getEntityGraph("person.name.with.city");
Map hints = new HashMap();
hints.put("javax.persistence.fetchgraph", graph);
return em.find(Person.class, personId, hints);
The same applies for queries, not only em.find method.
Look this tutorial for more details.
I think that that's not usual scenario of Data JPA usage. But you can achieve your goal with pure JPQL:
SELECT a.street, a.person.name FROM Address a WHERE …
This solution has 2 drawbacks:
It forces you to have bidirectional relationship Address ←→ Person
It returns List
Another solution (and that's preferred JPA way) is to create DTO like this:
class MyPersonDTO {
private String personName;
private List<String> cities;
public MyPersonDTO(String personName, List<Address> adresses) {
this.personName = personName;
cities = adresses
.stream()
.map(Address::getCity)
.collect(Collectors.toList());
}
}
And the execute JPQL query like this:
SELECT NEW package.MyPersonDTO(p.name, p.addressList) FROM Person p WHERE …
Return type will be List<MyPersonDTO> in that case.
Of course you can use any of this solutions inside #Query annotation and it should work.
Am new to Spring Boot & JPA...
Let's say I have two entities mapped to two tables which are joined in a database.
Student-1------<-Course
Also, lets presume that the database is already created and populated.
This depicts that one student has many courses...
My Student Entity:
#Entity
public class Student {
#OneToMany(mappedBy="student")
private List<Courses> courses;
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "Student_Id")
private long studentId;
#Column(name = "Student_Name")
private String studentName;
protected Student() { }
// Getters & Setters
}
My Course Entity:
#Entity
public class Course {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "Course_Id")
private long courseId;
#Id
#Column(name = "Student_Id")
private long studentId;
#ManyToOne
#PrimaryKeyJoinColumn(name="Student_Id", referencedColumnName="Student_Id")
private Student student;
#Column(name = "Course_Name")
private String courseName;
// Getters & Setters
}
In Spring Boot's Tutorial Guides, it illustrates how to extend a CrudRepository interface, but
it doesn't specify how to setup a Spring based DAO which contains custom finder methods which use HQL and EntityManager inside it.
Is the following DAO and DaoImpl correct?
public interface CourseDao {
List<Course> findCoursesByStudentName(String studentName);
}
#Repository
public class CourseDaoImpl implements CourseDao {
#PersistenceContext
EntityManager em;
public List<Course> findCoursesByStudentName(String studentName) {
String sql = "select c.courseName" +
"from Course c, Student s " +
"where c.course_id = s.student_id " +
"and s.studentName = :studentName ";
Query query = em.createQuery(sql);
query.setParameter("studentName", studentName);
return query.getResultList();
}
}
And then in the client code, for example, in the main class:
public class Application {
#Autowired
CustomerDao dao;
public static void main (String args []) {
List<Course> courses = dao.findCoursesByStudentName("John");
}
}
Is this the standard way to use HQL inside Spring DAOs ? I've seend examples of the #Transactional annotation being prepended to the DAO class's impl (e.g. CustomerDAOImpl) ?
Please let me know if this is the write way to structure my Spring Boot app or am I supposed to extend / add to the CrudRepository only?
If someone could correct my example and point me to a URL which talks about HQL using Entities that are joined, I would be very grateful.
The Spring Boot guides didn't depict joins or DAOs - I just need to learn how to properly create finder methods which emulate select statement which return lists or data structures.
Thanks for taking the time to read this...
If I understood your question correct you do have two questions:
How to create a DAO and DAOImpl?
Where to put your Transaction annotations?
In regards to the first question I want to point out that this is a question in regards to spring-data-jpa using Hibernate as a JPA provider, not spring-boot.
Using Spring Data I usually skip completely to create a DAO but directly use a Custom Repository extending a standard one like CrudRepository. So in your case you don't even have to write more code than:
#Repository
public interface StudentRepository extends CrudRepository<Student, Long> {
List<Student> findByStudentName(String studentName);
}
Which will be sufficient and Spring Data will take care of filling it with the correct implementation if you use
#Autowired
StudentRepository studentRepo;
in your service class. This is where I also usually annotate my methods with #Transactional to make sure that everything is working as expected.
In regards to your question about HQL please look into the spring data jpa documentation, which points out that for most of the cases it should be sufficient to stick to proper named methods in the interface or go for named queries (section 3.3.3) or use the #Query annotation (section 3.3.4) to manually define the query, e.g. should work (didn't tried):
#Repository
public interface #CourseRepository extends CrudRepository<Course, Long> {
#Query("select c.courseName from Course c, Student s where c.course_id = s.student_id and s.studentName = :studentName")
public List<Course> findCoursesByStudentName(String studentName);
}
If you annotate your CourseDaoImpl with #Transactional (Assuming your have defined JpaTransactionManager correctly) You can just retrieve the Student with the matching name and call the getCourses() method to lazy load the Courses attached to that student. Since findCoursesByStudentName will run within a Transaction it will load the courses just fine.
#Repository
#Transactional(readOnly=true)
public class CourseDaoImpl implements CourseDao {
#PersistenceContext
EntityManager em;
public List<Course> findCoursesByStudentName(String studentName) {
String sql = "select s " +
"from Student s " +
"where s.studentName = :studentName ";
Query query = em.createQuery(sql);
query.setParameter("studentName", studentName);
User user = query.getSingleResult();
if(user != null) {
return user.getCourses();
}
return new ArrayList<Course>();
}
}