JPA技术初认识

初次接触jpa写了一个小项目,下面是整个框架结构，很简单

1.首先把相应jar包导入到lib文件夹下
2.在根目录下创建一个“META-INF”文件夹，写配置文件persitence.xml
persitence.xml:

<?xml version="1.0" encoding="UTF-8"?><persistence version="2.0"    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_2_0.xsd">    <persistence-unit name="jpa" transaction-type="RESOURCE_LOCAL">        <!--         配置用什么ORM框架        1. 实际上配置的是 javax.persistence.spi.PersistenceProvider 接口的实现类        2. 如果JPA项目中只有一个JPA的实现产品，则可以不配置该节点         -->        <provider>org.hibernate.ejb.HibernatePersistence</provider>        <class>cn.lamb.jpa.User</class>        <properties>            <!-- 配置数据源信息 -->            <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>            <property name="javax.persistence.jdbc.url" value="jdbc:mysql://127.0.0.1:3306/jpa?charsetEncoding=utf8"/>            <property name="javax.persistence.jdbc.user" value="root"/>            <property name="javax.persistence.jdbc.password" value="ROOT"/>            <!-- 配置JPA实现产品的属性，即hibernate的属性 -->            <property name="hibernate.format_sql" value="true"/><!-- 是否格式化sql语句 -->            <property name="hibernate.show_sql" value="true"/> <!-- 是否在控制台打印sql语句 -->            <property name="hibernate.hbm2ddl.auto" value="update"/>        </properties>    </persistence-unit></persistence>

3.编写实体类
user.java

package cn.lamb.jpa;import javax.persistence.Column;import javax.persistence.Entity;import javax.persistence.GeneratedValue;import javax.persistence.GenerationType;import javax.persistence.Id;import javax.persistence.Table;@Table(name="USER")@Entitypublic class User {    private Integer id;    private String name;    private String email;    @Column(name="ID")    @GeneratedValue(strategy=GenerationType.AUTO)    @Id    public Integer getId() {        return id;    }    public void setId(Integer id) {        this.id = id;    }    @Column(name="NAME")    public String getName() {        return name;    }    public void setName(String name) {        this.name = name;    }    @Column(name="EMAIL")    public String getEmail() {        return email;    }    public void setEmail(String email) {        this.email = email;    }}

4.在数据库里创建一个数据库“jpa”,不用创建表
5.写一个测试类

package cn.lamb.jpa;import javax.persistence.EntityManager;import javax.persistence.EntityManagerFactory;import javax.persistence.EntityTransaction;import javax.persistence.Persistence;public class TEST {    public static void main(String[] args) {        //1. 创建EntityManagerFactory        String persistenceUnitName = "jpa";         EntityManagerFactory factory = Persistence.createEntityManagerFactory(persistenceUnitName);        //2. 创建EntityManager        EntityManager entityManager = factory.createEntityManager();        //3.开启事务        EntityTransaction transaction = entityManager.getTransaction();        transaction.begin();        //4. 持久化操作        User user = new User();        user.setName("tom2");        user.setEmail("[e-mail]");        //添加user到数据库，相当于hibernate的save();        entityManager.persist(user);        //5. 提交事务        transaction.commit();        //6. 关闭EntityManager        entityManager.close();        //7. 关闭EntityManagerFactory        factory.close();    }}

执行下测试类，看看吧！