Question

我有faces-config.xml：

<?xml version="1.0" encoding="UTF-8"?>
<faces-config
    xmlns="http://xmlns.jcp.org/xml/ns/javaee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-facesconfig_2_2.xsd"
    version="2.2">
    <managed-bean>
        <managed-bean-name>registerBean</managed-bean-name>
        <managed-bean-class>com.beans.RegisterBean</managed-bean-class>
        <managed-bean-scope>request</managed-bean-scope>
    </managed-bean>
</faces-config>

User课程：

package model;

import java.io.Serializable;
import javax.persistence.*;


/**
 * The persistent class for the users database table.
 * 
 */
@Entity
@Table(name="users")
@NamedQuery(name="User.findAll", query="SELECT u FROM User u")
public class User implements Serializable 
{
    private static final long serialVersionUID = 1L;

    @Id
    private int id;

    private String password;

    private String userName;

    public User() {
    }

    public int getId() {
        return this.id;
    }

    public void setId(int id) {
        this.id = id;
    }

    public String getPassword() {
        return this.password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    public String getUserName() {
        return this.userName;
    }

    public void setUserName(String userName) {
        this.userName = userName;
    }

}

和RegisterBean类：

package com.beans;

import javax.ejb.TransactionAttribute;
import javax.ejb.TransactionAttributeType;
import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import javax.persistence.PersistenceContextType;
import javax.transaction.UserTransaction;

import model.User;

//see faces-config.xml to see annotations

public class RegisterBean 
{
    @PersistenceContext(unitName="user-unit", type=PersistenceContextType.EXTENDED) 
    private EntityManager em;

    private String name;
    private String password;
    private String passwordRepeat;

    public String getPasswordRepeat() 
    {
        return passwordRepeat;
    }


    public void setPasswordRepeat(final String passwordRepeat) 
    {
        this.passwordRepeat = passwordRepeat;
    }


    public String getName ()
    {
        return name;
    }


    public void setName (final String name)
    {
        this.name = name;
    }


    public String getPassword ()
    {
        return password;
    }


    public void setPassword (final String password)
    {
        this.password = password;
    }

    public void register()
    {
        //TODO: validate username and password
        User newUser = new User();
        newUser.setPassword(password);
        newUser.setUserName(name);
        //User foundUser = em.find(User.class, 22); //this actually finds the user with Id 22

        em.persist(newUser); //TODO: find out why this doesn't work.
    }
}

问题：register()的{{1}}方法不会影响MySQL表格，即RegisterBean不会抛出任何异常，但实体不会进入我的数据库。任何人都可以帮助我吗？

编辑：如重新提出的那样的persistence.xml：

em.persist()

Answer 1

您正在JSF托管bean中使用扩展持久性上下文。这意味着您没有活动事务，并且所有DML都排队而不是执行。

see

如果与事务之外的扩展持久性上下文进行交互，则插入，更新和删除将排队，直到您访问事务中的持久性上下文。这意味着您调用的任何persist（），merge（）或remove（）方法实际上都不会导致JDBC执行，因此在您手动调用EntityManager.flush（）之前会更新数据库。

Answer 2

在Java EE 6中，为了创建事务，您需要EJB（容器管理事务）或自己管理事务边界（Bean管理事务）。请参阅链接的教程。

最简单的方法是两者中的第一个：实现它，只需在@Stateless EJB中移动相关代码（确保不扩展持久化上下文）

<强> RegisterBean

public class RegisterBean {    
    @EJB MyService service;

    public void register() {
        //TODO: validate username and password
        User newUser = new User();
        newUser.setPassword(password);
        newUser.setUserName(name);
        service.create(newUser);
    }
}

<强>为MyService

@Stateless
public class MyService {
    @PersistenceContext(unitName="user-unit") 
    private EntityManager em;

    public void create(User newUser) {
        em.persist(newUser);
    }
}

有用的链接

The Java EE Tutorial - Transactions

实体管理器不持久化类实例

2 个答案: