搭建JPA项目

　　环境: win7-X64, Eclipse, Hibernate5.0.2, MySQL

　　我们用Hibernate作为JPA的实现产品.

步骤:

1. 建立一个JAVA工程

2. 在 src 目录下面建立一个META_INF 目录, 然后在这个目录下面建立 persistence.xml 文件

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1"
    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_1.xsd">
</persistence>

 

3. 写配置文件 (persistence.xml)

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1"
    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_1.xsd">
    <!-- 
        name: 定义持久化单元的名字
        RESOURCE_LOCAL: 默认值，数据库级别的事务, 只能针对一种数据库, 不支持分布式事务
        JTA: 支持分布式事务
    -->
    <persistence-unit name="jpa-01" transaction-type="RESOURCE_LOCAL">
        <!-- 指定JPA的实现产品, 若项目中只有一个, 则可省略 -->
        <provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
        
        <!-- 实体类 -->
        <class>com.jpa.entities.Person</class>
        
        <!-- 连接数据库的信息 -->
        <properties>
            <property name="javax.persistence.jdbc.user" value="root"/>
            <property name="javax.persistence.jdbc.password" value="oracle"/>
            <property name="javax.persistence.jdbc.url" value="jdbc:mysql:///jpa"/>
            <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
            
            <property name="hibernate.hbm2ddl.auto" value="update"/>
            <property name="hibernate.show_sql" value="true"/>
            <property name="hibernate.format_sql" value="true"/>
        </properties>

    </persistence-unit>
</persistence>

 

4. 建立实体类

package com.jpa.entities;

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name = "PERSONS")
public class Person {

    @Id
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    private Integer id;
    private String name;
    private String email;

    public Integer getId() {
        return id;
    }

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

    public String getName() {
        return name;
    }

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

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

}

5. 建立测试类

package com.jpa.test;

import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.EntityTransaction;
import javax.persistence.Persistence;

import org.junit.After;
import org.junit.Before;
import org.junit.Test;

import com.jpa.entities.Person;

public class TestJPA {

    private EntityManagerFactory entityManagerFactory;
    private EntityManager entityManager;
    private EntityTransaction entityTransaction;

    @Before
    public void init() {
        entityManagerFactory = Persistence.createEntityManagerFactory("jpa-01");
        entityManager = entityManagerFactory.createEntityManager();
        entityTransaction = entityManager.getTransaction();
        entityTransaction.begin();
    }

    @After
    public void destory() {
        entityTransaction.commit();
        entityManager.close();
        entityManagerFactory.close();
    }

    @Test
    public void testConnection() {

        Person person = new Person();
        person.setName("Mike");
        person.setEmail("[email protected]");
        entityManager.persist(person);

    }

}

　　完成.