Reputation: 67
I am new in jpa and I am tryingg to write and execute a sample founded here:Sample JPA
Here is my persistance.xml file:
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" 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_0.xsd">
<persistence-unit name="EmployeeService" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<class>ir.ac.sbu.testsimplesql1.Employee</class>
<properties>
<property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/library?zeroDateTimeBehavior=convertToNull"/>
<property name="javax.persistence.jdbc.user" value="root"/>
<property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
<property name="javax.persistence.jdbc.password" value="pass"/>
<property name="hibernate.archive.autodetection" value="class"/>
<property name="hibernate.show_sql" value="true"/>
<property name="hibernate.format_sql" value="true"/>
<property name="hbm2ddl.auto" value="update"/>
</properties>
I have auto generate model class from database:
package ir.ac.sbu.testsimplesql1;
import java.io.Serializable;
import javax.persistence.Basic;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.NamedQueries;
import javax.persistence.NamedQuery;
import javax.persistence.Table;
import javax.xml.bind.annotation.XmlRootElement;
@Entity
@Table(name = "Employee")
@XmlRootElement
@NamedQueries({
@NamedQuery(name = "Employee.findAll", query = "SELECT e FROM Employee e"),
@NamedQuery(name = "Employee.findById", query = "SELECT e FROM Employee e WHERE e.id = :id"),
@NamedQuery(name = "Employee.findByFistName", query = "SELECT e FROM Employee e WHERE e.fistName = :fistName"),
@NamedQuery(name = "Employee.findByLastName", query = "SELECT e FROM Employee e WHERE e.lastName = :lastName"),
@NamedQuery(name = "Employee.findByDept", query = "SELECT e FROM Employee e WHERE e.dept = :dept")})
public class Employee implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@Basic(optional = false)
@Column(name = "id")
private Integer id;
@Column(name = "fistName")
private String fistName;
@Column(name = "lastName")
private String lastName;
@Column(name = "dept")
private String dept;
public Employee() {
}
public Employee(Integer id) {
this.id = id;
}
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public String getFistName() {
return fistName;
}
public void setFistName(String fistName) {
this.fistName = fistName;
}
public String getLastName() {
return lastName;
}
public void setLastName(String lastName) {
this.lastName = lastName;
}
public String getDept() {
return dept;
}
public void setDept(String dept) {
this.dept = dept;
}
@Override
public int hashCode() {
int hash = 0;
hash += (id != null ? id.hashCode() : 0);
return hash;
}
@Override
public boolean equals(Object object) {
// TODO: Warning - this method won't work in the case the id fields are not set
if (!(object instanceof Employee)) {
return false;
}
Employee other = (Employee) object;
if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id))) {
return false;
}
return true;
}
@Override
public String toString() {
return "ir.ac.sbu.testsimplesql1.Employee[ id=" + id + " ]";
}
}
This is my poem.xml: (Dependencies Tag)
<dependencies>
<dependency>
<groupId>org.eclipse.persistence</groupId>
<artifactId>javax.persistence</artifactId>
<version>2.0.0</version>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-entitymanager</artifactId>
<version>4.2.8.Final</version>
</dependency>
<dependency>
<groupId>mysql</groupId>
<artifactId>mysql-connector-java</artifactId>
<version>5.1.27</version>
</dependency>
<dependency>
<groupId>org.eclipse.persistence</groupId>
<artifactId>eclipselink</artifactId>
<version>2.5.2</version>
</dependency>
</dependencies>
And the last thing is my main function:
public class EmployeeTest {
private static EntityManager em;
public static void main(String[] args) {
EntityManagerFactory emf = Persistence.createEntityManagerFactory("EmployeeService");
em = emf.createEntityManager();
createEmployee(1, "Saint", "Peter", "Engineering");
createEmployee(2, "Jack", " Dorsey", "Imaginea");
createEmployee(3, "Sam", "Fox", "Imaginea");
}
private static void createEmployee(int id, String firstName, String lastName, String dept) {
em.getTransaction().begin();
Employee emp = new Employee();
emp.setId(id);
emp.setFistName(firstName);
emp.setLastName(lastName);
emp.setDept(dept);
em.persist(emp);
em.getTransaction().commit();
}
}
But when I run this code I face with this error:
I have searched this error bud couldn't understand what should I do. I've also visited this link: Can not find the declaration of element 'persistence'
javax.persistence.PersistenceException - JPA+Hibernate and alot more similar questions ....
Can anyone please help me to solve this error?
Thanks in advance for your attention.
Upvotes: 1
Views: 1682
Reputation: 341
You can check your XML against the XSD here : http://www.freeformatter.com/xml-validator-xsd.html Very useful.
Replace the beginning of your xml with:
<persistence 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_2_0.xsd"
version="2.0">
I don't know why, but yours is not validated.
Also, you miss the end of the xml, but i guess it's just a copy/paste error.
Upvotes: 3