我正在使用Java JPA
创建一个简单的应用程序,仅向表中插入一行数据(如果表不存在则创建它)。
我附上了一些代码作为可运行示例。
这是我得到的异常和堆栈跟踪:
EXCEPTION -- > org.hibernate.PersistentObjectException: detached entity passed to persist: view.Person
javax.persistence.PersistenceException: org.hibernate.PersistentObjectException: detached entity passed to persist: view.Person
at org.hibernate.jpa.spi.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1763)
at org.hibernate.jpa.spi.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1677)
at org.hibernate.jpa.spi.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1683)
at org.hibernate.jpa.spi.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:1187)
at view.TestJPA.main(TestJPA.java:34)
Caused by: org.hibernate.PersistentObjectException: detached entity passed to persist: view.Person
at org.hibernate.event.internal.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:139)
at org.hibernate.event.internal.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:75)
at org.hibernate.internal.SessionImpl.firePersist(SessionImpl.java:811)
at org.hibernate.internal.SessionImpl.persist(SessionImpl.java:784)
at org.hibernate.internal.SessionImpl.persist(SessionImpl.java:789)
at org.hibernate.jpa.spi.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:1181)
... 1 more
以下是我的代码:
主类:
package view;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.EntityTransaction;
import javax.persistence.Persistence;
public class TestJPA {
public static void main(String[] args) {
Person p = new Person(1, "Peter", "Parker");
EntityManagerFactory entityManagerFactory = Persistence.createEntityManagerFactory("TesePersistentUnit");
EntityManager entityManager = entityManagerFactory.createEntityManager();
EntityTransaction transaction = entityManager.getTransaction();
try {
transaction.begin();
entityManager.persist(p);
entityManager.getTransaction().commit();
}
catch (Exception e) {
if (transaction != null) {
transaction.rollback();
}
System.out.println("EXCEPTION -- > " + e.getMessage());
e.printStackTrace();
}
finally {
if (entityManager != null) {
entityManager.close();
}
}
}
}
而且Person类:
package view;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;
@Entity
@Table(name = "People")
public class Person {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private int id;
private String name;
private String lastName;
public Person(int id, String name, String lastName) {
this.id = id;
this.name = name;
this.lastName = lastName;
}
public Person() {
}
}
这是我的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-unit name="TesePersistentUnit" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<class>view.Person</class>
<properties>
<!-- SQL dialect -->
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect"/>
<property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/tese_tabelas?zeroDateTimeBehavior=convertToNull"/>
<property name="javax.persistence.jdbc.user" value="root"/>
<property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
<property name="javax.persistence.jdbc.password" value=""/>
<!-- Create/update tables automatically using mapping metadata -->
<property name="hibernate.hbm2ddl.auto" value="update"/>
</properties>
</persistence-unit>
</persistence>
----------------------- 编辑 ---------------------------
我刚刚将提供程序更改为 EclipseLink,而没有进行任何其他更改,它就可以工作了。我现在感到困惑。为什么使用 EclipseLink 可以工作,但使用 Hibernate 会生成异常?
EntityManager.merge
操作而不是EntityManager.persist
操作... - Carlitos Way