JPA로 JSF 애플리케이션을 빌드하려고합니다. 지금, 우리는 로그인 기능을 만들고 싶어하지만, 우리는 우리의 글래스 피쉬 서버에서 응용 프로그램을 실행할 때 예외가있다 :javax.persistence.PersistenceException를 해결하는 방법을 알 수 없습니다.
javax.persistence.PersistenceException :
우리는 생각 siteMami
라는 이름의 EntityManager 없음 지속성 제공 문제는 persistence.xml에서 어딘가에, 어쩌면 공급자에게, 제발 우리를 도와주세요. 감사!의 persistence.xml :
<?xml version="1.0" encoding="UTF-8"?>
<persistence xmlns="http://java.sun.com/xml/ns/persistence"
version="1.0">
<persistence-unit name="siteMami" transaction-type="JTA">
<provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
<class>model.Admin</class>
<class>model.User</class>
<class>model.Client</class>
<properties>
<property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver" />
<property name="javax.persistence.jdbc.url"
value="jdbc:mysql://localhost/siteMami" />
<property name="javax.persistence.jdbc.user" value="root" />
<property name="javax.persistence.jdbc.password" value="" />
</properties>
</persistence-unit>
</persistence>
User.java :
/**
*
*/
package model;
import java.io.Serializable;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.Inheritance;
import javax.persistence.InheritanceType;
import javax.persistence.Table;
import javax.persistence.Transient;
@Entity
@Table(name = "useri")
@Inheritance(strategy = InheritanceType.JOINED)
public class User implements Serializable
{
@Transient
private static long serialVersionUID = 6837935606727700935L;
@Id
@GeneratedValue
@Column(name = "idUseri")
private long id;
@Column(unique = true)
private String username;
private String password;
/**
* @param id
* @param userName
* @param password
*/
public User(long id, String username, String password)
{
super();
this.id = id;
this.username = username;
this.password = password;
}
/**
* @return the id
*/
public long getId()
{
return id;
}
/**
* @return the userName
*/
public String getUsername()
{
return username;
}
/**
* @return the password
*/
public String getPassword()
{
return password;
}
public void setId(long id)
{
this.id = id;
}
public void setUsername(String userName)
{
this.username = userName;
}
public void setPassword(String password)
{
this.password = password;
}
}
UserManager.java :
package dao;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.EntityTransaction;
import javax.persistence.Persistence;
import javax.persistence.PersistenceContext;
import javax.persistence.Query;
import model.User;
public class UserManager
{
private EntityManagerFactory factory;
public UserManager()
{
factory = Persistence.createEntityManagerFactory("siteMami");
}
public User getUser(String username, String password)
{
EntityManager entityManager = factory.createEntityManager();
EntityTransaction entityTransaction = entityManager.getTransaction();
entityTransaction.begin();
Query q = entityManager.createQuery("SELECT * FROM User WHERE User.username = '" + username + "' and User.password = '" + password + "'");
entityTransaction.commit();
return (User) q.getSingleResult();
}
}
Android Studio/ –