JPA:在独立的Java应用程序中保留后如何获取ID [英] JPA: How to get Id after persist in standalone java app

问题描述

这是一个独立的Java应用程序，而不是Web应用程序.所以当我坚持这样的对象

This is a standalone java application, not web application. So when I persist the object like this

public <T> T create(T t) {
    em.getTransaction().begin();
    em.persist(t);
    em.flush();
    em.getTransaction().commit();
    return t;
}

即使在数据库内部正确创建了具有正确数据和ID的新行，T t对象内部的id仍然为空.通常在使用@EJB的Web应用程序中，id在我持久化之后立即可用，因为它会将实体对象持久化到我的持久化上下文中，所以我不确定我是否在这里有我的持久化上下文?

The id inside the T t object is still null, even though a new row with correct data and id is created correctly inside the database. Usually in my web app that utilize @EJB, the id available right after I persist, since it persist the entity object into my persistence context, I am not sure if I have my persistence context here?

这就是我将ID映射到@Entity类中的方式

This is how I mapped my id inside my @Entity Class

@Id
@Basic(optional = false)
@Column(name = "ID")
private Long id;

我也像这样在数据库AUTO_INCREMENT中创建该表的ID

also I make the id of this table in the database AUTO_INCREMENT, like this

CREATE TABLE Config
(
    ID int NOT NULL AUTO_INCREMENT,
    PRIMARY KEY (ID)
)

这就是我获取EntityManager的方式

This is how I obtain my EntityManager

EntityManagerFactory emf = Persistence.createEntityManagerFactory("CorePU");
em = emf.createEntityManager();

这是我的persistence.xml

<persistence-unit name="CorePU" transaction-type="RESOURCE_LOCAL">
<provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>    
<class>com.wf.docsys.core.model.Config</class>    
<properties>
  <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/XNINFODB"/>
  <property name="javax.persistence.jdbc.password" value="xxx"/>
  <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
  <property name="javax.persistence.jdbc.user" value="xxx"/>
</properties>

请帮助，我不确定我在这里做错了什么.

Please help, I am not sure what I did wrong here.

推荐答案

这是我的答案.经过测试

映射您的ID时，请从上面的内容中切换

When mapping your Id, switch from what I have above, which is

@Id
@Basic(optional = false)
@Column(name = "ID")
private Long id;

到

@Entity
public class Person {
   @Id
   @TableGenerator(name="TABLE_GEN", table="SEQUENCE_TABLE", pkColumnName="SEQ_NAME",
       valueColumnName="SEQ_COUNT", pkColumnValue="PERSON_SEQ")
   @GeneratedValue(strategy=GenerationType.TABLE, generator="TABLE_GEN")
   private long id;
   ...
}

由于我使用Microsoft SQL Server，因此必须使用@TableGenerator.可在此处找到更多信息 http://en.wikibooks.org/wiki/Java_Persistence/Identity_and_Sequencing #Table_sequencing

Since I use microsoft SQL server, I have to use @TableGenerator. More information can be found here http://en.wikibooks.org/wiki/Java_Persistence/Identity_and_Sequencing#Table_sequencing

这篇关于JPA:在独立的Java应用程序中保留后如何获取ID的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！