我们正试图用JPA构建一个JSF应用程序。现在,我们要登录的功能,但是,当我们运行GlassFish服务器上的应用程序,有例外:无法弄清楚如何解决javax.persistence.PersistenceException
javax.persistence.PersistenceException:否EntityManager的持久性提供者名为siteMami
我们认为,问题来自persistence.xml,也许在提供商处,请帮助我们。谢谢!下面是目录结构:
的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();
}
}
如何在Android Studio中解决此问题/ – 2015-06-09 16:33:31