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我是JPA的新成员,我在Glassfish 3和Eclipse Java EE Web开发人员的JSF 2项目中使用。 这是我的设置和我想在我的数据库坚持方式:JPA不插入数据库,为什么?
的persistence.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="SuaParte" transaction-type="RESOURCE_LOCAL">
<class>com.suaparte.pojo.Area</class>
//other entities
<properties>
<property name="eclipselink.jdbc.batch-writing" value="JDBC"/>
<property name="javax.persistence.jdbc.url" value="jdbc:mysql://<hostname>:3306/sua_parte"/>
<property name="javax.persistence.jdbc.user" value=<username>/>
<property name="javax.persistence.jdbc.password" value=<password>/>
<property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
</properties>
</persistence-unit>
</persistence>
我的实体:
@Entity
@Table(name="area")
public class Area implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(unique=true, nullable=false)
private byte id;
@Column(nullable=false, length=45)
private String area;
//bi-directional many-to-one association to Company
@OneToMany(mappedBy="areaBean")
private List<Company> companies;
//getters and setters
}
怎么我打电话我EntityManager并尝试坚持对象:
public static void main(String[] args) {
// TODO Auto-generated method stub
EntityManagerFactory entityManagerFactory = Persistence.createEntityManagerFactory("SuaParte");
EntityManager entityManager = entityManagerFactory.createEntityManager();
Area area = new Area();
area.setArea("test");
entityManager.persist(area);
}
但是,当我什么都不执行发生在我的数据库中,JPA不会将该对象保留在我的表中,我做错了什么? 有什么想法?
感谢兄弟=),它的工作,最后。 –
如果您最终在JSF中实现它,请在EJB中执行它,然后由容器为您处理事务。 – BalusC