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Hibernate doesn't creating table based on entity

After starting the program (launching TomCat) there are no tables created in the schema, but the table "player" has to be created automatically.
I checked hibernate config, but can't find where is the problem.
I've tried changing hbm2ddl.auto to hibernate.hbm2ddl.auto (also create, create-drop etc.) but it didn't help.
If there are any ideas, please let me know. Thanks.
Entity class:
package com.game.entity;
import javax.persistence.*;
import java.util.Date;
#Entity
#Table(schema = "rpg", name = "player")
public class Player {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id")
private Long id;
#Column(name = "name", length = 12, nullable = false)
private String name;
#Column(name = "title", length = 30, nullable = false)
private String title;
#Column(name = "race", nullable = false)
#Enumerated(EnumType.ORDINAL)
private Race race;
#Column(name = "profession", nullable = false)
#Enumerated(EnumType.ORDINAL)
private Profession profession;
#Column(name = "birthday", nullable = false)
private Date birthday;
#Column(name = "banned", nullable = false)
private Boolean banned;
#Column(name = "level", nullable = false)
private Integer level;
public Player() {
}
public Player(Long id, String name, String title, Race race, Profession profession, Date birthday, Boolean banned, Integer level) {
this.id = id;
this.name = name;
this.title = title;
this.race = race;
this.profession = profession;
this.birthday = birthday;
this.banned = banned;
this.level = level;
}
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getTitle() {
return title;
}
public void setTitle(String title) {
this.title = title;
}
public Race getRace() {
return race;
}
public void setRace(Race race) {
this.race = race;
}
public Profession getProfession() {
return profession;
}
public void setProfession(Profession profession) {
this.profession = profession;
}
public Date getBirthday() {
return birthday;
}
public void setBirthday(Date birthday) {
this.birthday = birthday;
}
public Boolean getBanned() {
return banned;
}
public void setBanned(Boolean banned) {
this.banned = banned;
}
public Integer getLevel() {
return level;
}
public void setLevel(Integer level) {
this.level = level;
}
}
Repository class:
package com.game.repository;
import com.game.entity.Player;
import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.boot.registry.StandardServiceRegistry;
import org.hibernate.boot.registry.StandardServiceRegistryBuilder;
import org.hibernate.cfg.Configuration;
import org.hibernate.query.NativeQuery;
import org.springframework.stereotype.Repository;
import javax.annotation.PreDestroy;
import java.util.List;
import java.util.Optional;
#Repository(value = "db")
public class PlayerRepositoryDB implements IPlayerRepository {
private final SessionFactory sessionFactory;
public PlayerRepositoryDB() {
Configuration configuration = new Configuration().configure().addAnnotatedClass(Player.class);
StandardServiceRegistry serviceRegistry = new StandardServiceRegistryBuilder()
.applySettings(configuration.getProperties()).build();
sessionFactory = configuration.buildSessionFactory(serviceRegistry);
}
#Override
public List<Player> getAll(int pageNumber, int pageSize) {
try(Session session = sessionFactory.openSession()){
NativeQuery<Player> nativeQuery = session.createNativeQuery("SELECT * FROM rpg.player", Player.class);
nativeQuery.setFirstResult(pageNumber * pageSize);
nativeQuery.setMaxResults(pageSize);
return nativeQuery.list();
}
}
Hibernate configuration:
<?xml version='1.0' encoding='utf-8'?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD//EN"
"http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
<session-factory>
<property name="connection.url">jdbc:mysql://localhost:3306/rpg</property>
<property name="connection.driver_class">com.mysql.cj.jdbc.Driver</property>
<property name="connection.username">root</property>
<property name="connection.password">1234</property>
<property name="hbm2ddl.auto">update</property>
<property name="dialect">org.hibernate.dialect.MySQL8Dialect</property>
<property name="show_sql">true</property>
<property name="hibernate.current_session_context_class">thread</property>
</session-factory>
</hibernate-configuration>
Full project code with pom.xml is available by link:
https://github.com/gamlethot/project-hibernate-1

1-Hibernate does not recognize your repository. You should not mark repo classes as #Repository because they are not interfaces and in your example they are working like a service. So they can be #Service.
2-Do not implement IPlayerRepository. Mark it as #Repository and just autowire it to your service classes (or use constructor injection and just use like a variable)
Like:
#Service
public class PlayerRepositoryDB {
private IPlayerRepository playerRepository;
public PlayerRepositoryDB (IPlayerRepository playerRepository){ //CONSTRUCTOR
this.playerRepository = playerRepository;...
3- DB repository classes are implementing IPlayerRepository but it must be marked as #Repository and It should extend either CrudRepository or JpaRepository (which extends CrudRepository already).
Like:
#Repository
public interface IPlayerRepository extends JpaRepository<Player, Long> {
//Here are the methods;
}
Here, the Long is the type of primary key of Player class.

Hibernate XML:
<property name="hibernate.connection.CharSet">utf8mb4</property>
<property name="hibernate.connection.characterEncoding">UTF-8</property>
<property name="hibernate.connection.useUnicode">true</property>
Connection url:
db.url=jdbc:mysql://localhost:3306/db_name?useUnicode=true&character_set_server=utf8mb4
As a side note I would like to make one clarification that UTF-8 is the character encoding while utf8mb4 is a character set that MySQL supports. MySQL's utf8mb4 is a superset to MySQL's utf8.
Spring/Hibernate filter:
<form accept-charset="UTF-8">

Problem solved.
It was because of javax.persistence.; import instead of
jakarta.persistence. in entity class.

Error with JPA persistence.xml No Persistence provider for EntityManager

I have pom.xml
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/maven-v4_0_0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>net.vyhn</groupId>
<artifactId>jpa</artifactId>
<packaging>war</packaging>
<version>1.0-SNAPSHOT</version>
<name>jpa Maven Webapp</name>
<url>http://maven.apache.org</url>
<dependencies>
<dependency>
<groupId>junit</groupId>
<artifactId>junit</artifactId>
<version>4.12</version>
<scope>test</scope>
</dependency>
<dependency>
<groupId>org.eclipse.persistence</groupId>
<artifactId>eclipselink</artifactId>
<version>2.6.3</version>
</dependency>
<dependency>
<groupId>org.eclipse.persistence</groupId>
<artifactId>org.eclipse.persistence.core</artifactId>
<version>2.6.3</version>
</dependency>
<dependency>
<groupId>org.eclipse.persistence</groupId>
<artifactId>org.eclipse.persistence.jpa</artifactId>
<version>2.6.3</version>
</dependency>
<dependency>
<groupId>javax.persistence</groupId>
<artifactId>persistence-api</artifactId>
<version>1.0.2</version>
</dependency>
<dependency>
<groupId>mysql</groupId>
<artifactId>mysql-connector-java</artifactId>
<version>6.0.2</version>
</dependency>
</dependencies>
<build>
<finalName>jpa</finalName>
</build>
</project>
Employee.java
package example.com.domain;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.ManyToOne;
#Entity
public class Employee {
#Id
#GeneratedValue
private Long id;
private String name;
#ManyToOne
private Department department;
public Employee() {
}
public Employee(String name, Department department) {
this.name = name;
this.department = department;
}
public Employee(String name) {
this.name = name;
}
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public Department getDepartment() {
return department;
}
public void setDepartment(Department department) {
this.department = department;
}
#Override
public String toString() {
return "Employee [id=" + id + ", name=" + name + ", department="
+ department.getName() + "]";
}
}
Department.java
package example.com.domain;
import javax.persistence.CascadeType;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.OneToMany;
import java.util.ArrayList;
import java.util.List;
#Entity
public class Department {
#Id
#GeneratedValue
private Long id;
private String name;
#OneToMany(mappedBy = "department", cascade = CascadeType.PERSIST)
private List<Employee> employees = new ArrayList<Employee>();
public Department() {
super();
}
public Department(String name) {
this.name = name;
}
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public List<Employee> getEmployees() {
return employees;
}
public void setEmployees(List<Employee> employees) {
this.employees = employees;
}
}
JpaTest.java
package example.com.jpa;
import example.com.domain.Department;
import example.com.domain.Employee;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.EntityTransaction;
import javax.persistence.Persistence;
import java.util.List;
public class JpaTest {
private EntityManager manager;
public JpaTest(EntityManager manager) {
this.manager = manager;
}
public static void main(String[] args) {
EntityManagerFactory factory = Persistence.createEntityManagerFactory("persistenceUnit");
EntityManager manager = factory.createEntityManager();
JpaTest test = new JpaTest(manager);
EntityTransaction tx = manager.getTransaction();
tx.begin();
try {
test.createEmployees();
} catch (Exception e) {
e.printStackTrace();
}
tx.commit();
test.listEmployees();
System.out.println(".. done");
}
private void createEmployees() {
int numOfEmployees = manager.createQuery("Select a From Employee a", Employee.class).getResultList().size();
if (numOfEmployees == 0) {
Department department = new Department("java");
manager.persist(department);
manager.persist(new Employee("Jakab Gipsz", department));
manager.persist(new Employee("Captain Nemo", department));
}
}
private void listEmployees() {
List<Employee> resultList = manager.createQuery("Select a From Employee a", Employee.class).getResultList();
System.out.println("num of employess:" + resultList.size());
for (Employee next : resultList) {
System.out.println("next employee: " + next);
}
}
}
When I run JpaTest.java, has error:
"C:\Program Files\Java\jdk1.8.0_92\bin\java" -Didea.launcher.port=7533 "-Didea.launcher.bin.path=C:\Program Files (x86)\JetBrains\IntelliJ IDEA 2016.1.2\bin" -Dfile.encoding=UTF-8 -classpath "C:\Program Files\Java\jdk1.8.0_92\jre\lib\charsets.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\deploy.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\ext\access-bridge-64.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\ext\cldrdata.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\ext\dnsns.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\ext\jaccess.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\ext\jfxrt.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\ext\localedata.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\ext\nashorn.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\ext\sunec.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\ext\sunjce_provider.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\ext\sunmscapi.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\ext\sunpkcs11.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\ext\zipfs.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\javaws.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\jce.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\jfr.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\jfxswt.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\jsse.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\management-agent.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\plugin.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\resources.jar;C:\Program Files\Java\jdk1.8.0_92\jre\lib\rt.jar;C:\Users\Administrator\IdeaProjects\demo_jpa\target\classes;C:\Users\Administrator\.m2\repository\org\eclipse\persistence\eclipselink\2.6.3\eclipselink-2.6.3.jar;C:\Users\Administrator\.m2\repository\org\eclipse\persistence\javax.persistence\2.1.1\javax.persistence-2.1.1.jar;C:\Users\Administrator\.m2\repository\org\eclipse\persistence\commonj.sdo\2.1.1\commonj.sdo-2.1.1.jar;C:\Users\Administrator\.m2\repository\javax\validation\validation-api\1.1.0.Final\validation-api-1.1.0.Final.jar;C:\Users\Administrator\.m2\repository\org\glassfish\javax.json\1.0.4\javax.json-1.0.4.jar;C:\Users\Administrator\.m2\repository\org\eclipse\persistence\org.eclipse.persistence.core\2.6.3\org.eclipse.persistence.core-2.6.3.jar;C:\Users\Administrator\.m2\repository\org\eclipse\persistence\org.eclipse.persistence.asm\2.6.3\org.eclipse.persistence.asm-2.6.3.jar;C:\Users\Administrator\.m2\repository\org\eclipse\persistence\org.eclipse.persistence.jpa\2.6.3\org.eclipse.persistence.jpa-2.6.3.jar;C:\Users\Administrator\.m2\repository\org\eclipse\persistence\org.eclipse.persistence.antlr\2.6.3\org.eclipse.persistence.antlr-2.6.3.jar;C:\Users\Administrator\.m2\repository\org\eclipse\persistence\org.eclipse.persistence.jpa.jpql\2.6.3\org.eclipse.persistence.jpa.jpql-2.6.3.jar;C:\Users\Administrator\.m2\repository\javax\persistence\persistence-api\1.0.2\persistence-api-1.0.2.jar;C:\Users\Administrator\.m2\repository\mysql\mysql-connector-java\6.0.2\mysql-connector-java-6.0.2.jar;C:\Program Files (x86)\JetBrains\IntelliJ IDEA 2016.1.2\lib\idea_rt.jar" com.intellij.rt.execution.application.AppMain example.com.jpa.JpaTest
Exception in thread "main" javax.persistence.PersistenceException: No Persistence provider for EntityManager named persistenceUnit
at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:85)
at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:54)
at example.com.jpa.JpaTest.main(JpaTest.java:20)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:498)
at com.intellij.rt.execution.application.AppMain.main(AppMain.java:144)
Help me fix this error, thank you!

You have to put persistence.xml in java/resources folder.
<persistence-unit name="persistenceUnit" transaction-type="RESOURCE_LOCAL">
<provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
I hope this will work

Setting up JPA persistence with exception No persistence provider for EntityManager

I am overlooking some aspect of the setup but I don't know where to look. Eclipse, Tomcat, and MySQL. In the servlet I have an exception: No Persistence provider for EntityManager named EmployeeService. MySQL server is running and I configured a connection for the project. I have eclipselink jar library in the build path. What can I check to get it working?
Employee.java
package servlet;
import java.io.Serializable;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.Table;
#Entity
#Table(name="employee")
public class Employee implements Serializable {
private static final long serialVersionUID = 1L;
#Id
private int id;
private String name;
private double salary;
public Employee() { }
public Employee(int id){ this.id = id; }
public int getId() { return this.id; }
public void setId(int id) { this.id = id; }
public String getName(){ return this.name; }
public void setName(String name){ this.name = name; }
public double getSalary(){ return salary; }
public void setSalary(double salary){ this.salary = salary; }
public String toString(){ return "Employee: " + name; }
}
EmployeeService.java
package servlet;
import java.util.List;
import javax.persistence.EntityManager;
import javax.persistence.TypedQuery;
public class EmployeeService {
protected EntityManager em;
public EmployeeService(EntityManager em){
this.em = em;
}
public Employee createEmployee(int id, String name, Double salary){
Employee emp = new Employee(id);
emp.setName(name);
emp.setSalary(salary);
em.persist(emp);
return emp;
}
public List<Employee> findAllEmployees(){
TypedQuery<Employee> query = em.createQuery("SELECT e FROM Employee e", Employee.class);
return query.getResultList();
}
}
Relevant servlet code
protected void processRequest(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
EntityManagerFactory emf = Persistence.createEntityManagerFactory("EmployeeService");
EntityManager em = emf.createEntityManager();
EmployeeService service = new EmployeeService(em);
String name = request.getParameter("name");
em.getTransaction().begin();
Employee emp = service.createEmployee(158, name, 279445.00);
em.getTransaction().commit();
System.out.println(emp);
List<Employee> list = service.findAllEmployees();
for(Employee e : list){
System.out.println(e);
}
}
persistence.xml
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
<persistence-unit name="4_WEB_JPA" transaction-type="RESOURCE_LOCAL">
<provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
<class>servlet.Employee</class>
<properties>
<property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
<property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/company"/>
<property name="javax.persistence.jdbc.user" value="root"/>
<property name="javax.persistence.jdbc.password" value=""/>
</properties>
</persistence-unit>
</persistence>

You can initiate EntityManagerFactory by
EntityManagerFactory emf = Persistence.createEntityManagerFactory("4_WEB_JPA");
because your persistent unit named 4_WEB_JPA in your persistent.xml.

How can I change this configuration so I can work with Hibernate?

This is my project structure:
I have these following files:
context.xml
<?xml version='1.0' encoding='utf-8'?>
<Context>
<Resource name="jdbc/sakila"
url="jdbc:mysql://localhost/sakila"
username="root"
password=""
driverClassName="com.mysql.jdbc.Driver"
type="javax.sql.DataSource"
maxActive="20" maxIdle="5" maxWait="10000"
/>
</Context>
persistence.xml
<?xml version="1.0" encoding="UTF-8"?>
<persistence xmlns="http://xmlns.jcp.org/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence
http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd"
version="2.1">
<persistence-unit name="sakila" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
<non-jta-data-source>java:comp/env/jdbc/sakila</non-jta-data-source>
</persistence-unit>
</persistence>
And in a DAO class I can use:
EntityManagerFactory sakilaFactory = Persistence.createEntityManagerFactory("sakila");
EntityManager entityManager = sakilaFactory.createEntityManager();
which all works fine..
But I want to change my configuration to using .xml files and instead of EntityManager I want to use Session from Hibernate API.
I have looked at several tutorials on the web but I am only more confused. How should I use the .xml files for Hibernate configuration?
I am running the application on Tomcat and here is my dependencies from pom.xml
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-core</artifactId>
<version>4.3.10.Final</version>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-entitymanager</artifactId>
<version>4.3.10.Final</version>
</dependency>
<dependency>
<groupId>mysql</groupId>
<artifactId>mysql-connector-java</artifactId>
<version>5.1.6</version>
<scope>provided</scope>
</dependency>
What I want to achieve is a proper Hibernate configuration, and using Session instead of EntityManager.
Edit #1: This is what my Actor class looks like:
package biz.tugay.model;
/* User: koray#tugay.biz Date: 06/08/15 Time: 09:17 */
import javax.persistence.*;
import java.sql.Timestamp;
#Entity
#Table(name = "actor")
public class Actor {
private Integer actor_id;
private String first_name;
private String last_name;
private Timestamp last_update;
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
public Integer getActor_id() {
return actor_id;
}
public void setActor_id(Integer id) {
this.actor_id = id;
}
public String getFirst_name() {
return first_name;
}
public void setFirst_name(String first_name) {
this.first_name = first_name;
}
public String getLast_name() {
return last_name;
}
public void setLast_name(String last_name) {
this.last_name = last_name;
}
public Timestamp getLast_update() {
return last_update;
}
public void setLast_update(Timestamp last_update) {
this.last_update = last_update;
}
#Override
#Transient
public String toString() {
return "Actor{" +
"id=" + actor_id +
", first_name='" + first_name + '\'' +
", last_name='" + last_name + '\'' +
", last_update=" + last_update +
'}';
}
}

<hibernate-configuration>
<session-factory>
<property name="hibernate.connection.driver_class">com.mysql.jdbc.Driver</property>
<property name="hibernate.connection.url">jdbc:mysql://localhost:PORT/databasename</property>
<property name="hibernate.connection.username">username</property>
<property name="hibernate.connection.password">XXXX</property>
<property name="hibernate.dialect">org.hibernate.dialect.MySQLDialect</property>
<property name="show_sql">true</property>
<property name="format_sql">true</property>
<property name="hbm2ddl.auto">create </property>
<property name="hbm2ddl.auto">validate</property>
<mapping class="com.packageClass"/>
<mapping class="com.package.OtherClass"/>
</session-factory>
you should have the hibernate.cfg.xml - configuration file for hibernate; i gave you a template above.It should be placed in the src of your java resources. Now you can use annotations for persistance. And you should also add #Column annotation on every attribute in you actor class to map every attribute on an attribute from your table.Here is an example class
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;
#Entity
#Table(name="useri")
public class User implements Serializable {
#Id
#GeneratedValue(strategy=GenerationType.AUTO)
private Integer id_user;
#Column(name="username")
private String username;
#Column(name="password")
private String password;
#Column(name="rol")
private String rol;
public Integer getId_user() {
return id_user;
}
public void setId_user(Integer id_user) {
this.id_user = id_user;
}
public String getUsername() {
return username;
}
public void setUsername(String username) {
this.username = username;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public String getRol() {
return rol;
}
public void setRol(String rol) {
this.rol = rol;
}
public User() {
}
}

The target entity of the relationship attribute cannot be determined

I can't figure out what the problem is, please help. I've searched a lot, but didn't find any useful advice. Maybe you can help me. Thanks a lot.
There are two classes using eclipselink as jpa provider:
#Entity
#Table(name = "USER")
public class User {
#Id
#GeneratedValue(strategy = GenerationType.TABLE)
private Long id;
private String login;
private Long groupId;
private String email;
#ManyToMany(mappedBy = "users")
private List polls;
#OneToMany(mappedBy = "user")
private List votes;
public List getVotes() {
return votes;
}
public void setVotes(List votes) {
this.votes = votes;
}
public User() {
}
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getLogin() {
return login;
}
public void setLogin(String login) {
this.login = login;
}
public Long getGroupId() {
return groupId;
}
public void setGroupId(Long groupId) {
this.groupId = groupId;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public List getPolls() {
return polls;
}
public void setPolls(List polls) {
this.polls = polls;
}
}
and
#Entity
#Table(name = "VOTE")
public class Vote {
#Id
#GeneratedValue(strategy = GenerationType.TABLE)
private Long vid;
private String comment;
#ManyToOne
private User user;
#ManyToOne
private Option option;
public Vote() {
}
public Long getVid() {
return vid;
}
public void setVid(Long vid) {
this.vid = vid;
}
#Column(name = "comment")
public String getComment() {
return comment;
}
public void setComment(String comment) {
this.comment = comment;
}
public User getUser() {
return user;
}
public void setUser(User user) {
this.user = user;
}
public Option getOption() {
return option;
}
public void setOption(Option option) {
this.option = option;
}
}
When I'm trying to compile this, I receive error:
Exception [EclipseLink-7214] (Eclipse Persistence Services - 2.4.0.v20120608-r11652): org.eclipse.persistence.exceptions.ValidationException
Exception Description: The target entity of the relationship attribute [votes] on the class [class logic.User] cannot be determined. When not using generics, ensure the target entity is defined on the relationship mapping.
Here is my persistence.xml file:
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="1.0"
xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd">
<persistence-unit name="pollsPU" transaction-type="RESOURCE_LOCAL">
<provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
<class>logic.Option</class>
<class>logic.Poll</class>
<class>logic.User</class>
<class>logic.Vote</class>
<class>logic.Greeting</class>
<properties>
<property name="eclipselink.jdbc.password" value="" />
<property name="eclipselink.jdbc.user" value="" />
<property name="eclipselink.jdbc.driver" value="com.mysql.jdbc.Driver" />
<property name="eclipselink.jdbc.url"
value="jdbc:mysql://localhost:3306/core_polls" />
<property name="eclipselink.ddl-generation" value="drop-and-create-tables" />
<property name="eclipselink.logging.level" value="INFO" />
<property name="eclipselink.ddl-generation.output-mode"
value="database" />
</properties>
</persistence-unit>
</persistence>

As stated in the Javadoc
If the collection is defined using generics to specify the element
type, the associated target entity type need not be specified;
otherwise the target entity class must be specified.
So, you can do either
#OneToMany(mappedBy = "user")
private List<Vote> votes;
or
#OneToMany(targetEntity=logic.Vote.class, mappedBy = "user")
private List votes;
But I would prefer the first.