在完成之前2篇文章内容之后，我们讨论下以下情况。
项目要改版，TUser或者User要修改属性，添加或者删除某些字段，那么数据库已保存的数据还能被反序列化为原有对象，并进行扩展么？
修改User.java
package cn.vicky.po.vo;
import java.io.Serializable;
/**
*
* @author Vicky.H
*/
public class User implements Serializable {
private static final long serialVersionUID = 1L;
/** 新增加的属性 **/
private String address;
private String username;
private String password;
private boolean sex;
private short age;
/** 新增加的属性 **/
private int cardnum;
public String getAddress() {
return address;
}
public void setAddress(String address) {
this.address = address;
}
public short getAge() {
return age;
}
public void setAge(short age) {
this.age = age;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public boolean isSex() {
return sex;
}
public void setSex(boolean sex) {
this.sex = sex;
}
public String getUsername() {
return username;
}
public void setUsername(String username) {
this.username = username;
}
public int getCardnum() {
return cardnum;
}
public void setCardnum(int cardnum) {
this.cardnum = cardnum;
}
}
修改App.java
package cn.vicky.tmp1;
import cn.vicky.po.TUser;
import cn.vicky.po.vo.User;
import java.io.IOException;
import java.sql.SQLException;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;
/**
*
*/
public class App {
public static void create1(EntityManager em) {
User user = new User();
user.setUsername("Vicky");
user.setPassword("123456");
short age = 25;
user.setAge(age);
user.setSex(true);
TUser tUser = new TUser();
tUser.setUser(user);
em.persist(tUser);
}
public static void create2(EntityManager em) {
User user = new User();
user.setUsername("Jack");
user.setPassword("654321");
short age = 37;
user.setAge(age);
user.setSex(true);
TUser tUser = new TUser();
tUser.setUser(user);
em.persist(tUser);
}
/**
* 通过主键获得TUser对象，并且获得User对象
* @param em
*/
public static void find1(EntityManager em) {
TUser tUser = em.find(TUser.class, 1);
User user = tUser.getUser();
System.out.println("user.getUsername() = " + user.getUsername());
System.out.println("user.getPassword() = " + user.getPassword());
System.out.println("user.getAge() = " + user.getAge());
System.out.println("user.isSex() = " + user.isSex());
System.out.println("user.getAddress() = " + user.getAddress());
System.out.println("user.getCardnum() = " + user.getCardnum());
}
/**
* 直接查询Blob数据并强制转换为User对象
* @param em
*/
public static void find2(EntityManager em) {
User user = (User) em.createQuery("select o.user from t_user o where o.id = 1").getSingleResult();
System.out.println("user.getUsername() = " + user.getUsername());
System.out.println("user.getPassword() = " + user.getPassword());
System.out.println("user.getAge() = " + user.getAge());
System.out.println("user.isSex() = " + user.isSex());
System.out.println("user.getAddress() = " + user.getAddress());
System.out.println("user.getCardnum() = " + user.getCardnum());
}
public static void main(String[] args) throws SQLException, IOException, ClassNotFoundException {
EntityManagerFactory emf = Persistence.createEntityManagerFactory("pu");
EntityManager em = emf.createEntityManager();
em.getTransaction().begin();
// create1(em);
// create2(em);
find1(em);
find2(em);
em.getTransaction().commit();
em.close();
emf.close();
}
}
运行：
Hibernate: select tuser0_.id as id0_0_, tuser0_.user as user0_0_ from t_user tuser0_ where tuser0_.id=?
user.getUsername() = Vicky
user.getPassword() = 123456
user.getAge() = 25
user.isSex() = true
user.getAddress() = null
user.getCardnum() = 0
Hibernate: select tuser0_.user as col_0_0_ from t_user tuser0_ where tuser0_.id=1 limit ?
user.getUsername() = Vicky
user.getPassword() = 123456
user.getAge() = 25
user.isSex() = true
user.getAddress() = null
user.getCardnum() = 0
尝试删除属性后，程序执行依然正确。似乎挺简单的，没有什么值得注意的。
但这里的确有个地方需要注意，那就是：
private static final long serialVersionUID = 1L;
我们尝试删除User.java 的 private static final long serialVersionUID = 1L;
再次运行App.java
Hibernate: select tuser0_.id as id0_0_, tuser0_.user as user0_0_ from t_user tuser0_ where tuser0_.id=?
Exception in thread "main" javax.persistence.PersistenceException: org.hibernate.type.SerializationException: could not deserialize
at org.hibernate.ejb.AbstractEntityManagerImpl.throwPersistenceException(AbstractEntityManagerImpl.java:614)
at org.hibernate.ejb.AbstractEntityManagerImpl.find(AbstractEntityManagerImpl.java:202)
at cn.vicky.tmp1.App.find1(App.java:47)
at cn.vicky.tmp1.App.main(App.java:81)
Caused by: org.hibernate.type.SerializationException: could not deserialize
at org.hibernate.util.SerializationHelper.deserialize(SerializationHelper.java:188)
at org.hibernate.util.SerializationHelper.deserialize(SerializationHelper.java:211)
at org.hibernate.type.SerializableType.fromBytes(SerializableType.java:105)
at org.hibernate.type.SerializableType.get(SerializableType.java:62)
at org.hibernate.type.NullableType.nullSafeGet(NullableType.java:184)
at org.hibernate.type.NullableType.nullSafeGet(NullableType.java:173)
at org.hibernate.type.AbstractType.hydrate(AbstractType.java:105)
at org.hibernate.persister.entity.AbstractEntityPersister.hydrate(AbstractEntityPersister.java:2124)
at org.hibernate.loader.Loader.loadFromResultSet(Loader.java:1404)
at org.hibernate.loader.Loader.instanceNotYetLoaded(Loader.java:1332)
at org.hibernate.loader.Loader.getRow(Loader.java:1230)
at org.hibernate.loader.Loader.getRowFromResultSet(Loader.java:603)
at org.hibernate.loader.Loader.doQuery(Loader.java:724)
at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:259)
at org.hibernate.loader.Loader.loadEntity(Loader.java:1881)
at org.hibernate.loader.entity.AbstractEntityLoader.load(AbstractEntityLoader.java:71)
at org.hibernate.loader.entity.AbstractEntityLoader.load(AbstractEntityLoader.java:65)
at org.hibernate.persister.entity.AbstractEntityPersister.load(AbstractEntityPersister.java:3072)
at org.hibernate.event.def.DefaultLoadEventListener.loadFromDatasource(DefaultLoadEventListener.java:434)
at org.hibernate.event.def.DefaultLoadEventListener.doLoad(DefaultLoadEventListener.java:415)
at org.hibernate.event.def.DefaultLoadEventListener.load(DefaultLoadEventListener.java:165)
at org.hibernate.event.def.DefaultLoadEventListener.proxyOrLoad(DefaultLoadEventListener.java:223)
at org.hibernate.event.def.DefaultLoadEventListener.onLoad(DefaultLoadEventListener.java:126)
at org.hibernate.impl.SessionImpl.fireLoad(SessionImpl.java:905)
at org.hibernate.impl.SessionImpl.get(SessionImpl.java:842)
at org.hibernate.impl.SessionImpl.get(SessionImpl.java:835)
at org.hibernate.ejb.AbstractEntityManagerImpl.find(AbstractEntityManagerImpl.java:182)
... 2 more
Caused by: java.io.InvalidClassException: cn.vicky.po.vo.User; local class incompatible: stream classdesc serialVersionUID = 1, local class serialVersionUID = 993454148924749307
at java.io.ObjectStreamClass.initNonProxy(ObjectStreamClass.java:562)
at java.io.ObjectInputStream.readNonProxyDesc(ObjectInputStream.java:1583)
at java.io.ObjectInputStream.readClassDesc(ObjectInputStream.java:1496)
at java.io.ObjectInputStream.readOrdinaryObject(ObjectInputStream.java:1732)
at java.io.ObjectInputStream.readObject0(ObjectInputStream.java:1329)
at java.io.ObjectInputStream.readObject(ObjectInputStream.java:351)
at org.hibernate.util.SerializationHelper.deserialize(SerializationHelper.java:181)
... 28 more
错误原因:
serialVersionUID用来表明类的不同版本间的兼容性 简单来说，Java的序列化机制是通过在运行时判断类的serialVersionUID来验证版本一致性的。在进行反序列化时，JVM会把传来 的字节流中的serialVersionUID与本地相应实体（类）的serialVersionUID进行比较，如果相同就认为是一致的，可以进行反序 列化，否则就会出现序列化版本不一致的异常。
当实现java.io.Serializable接口的实体（类）没有显式地定义一个名为serialVersionUID，类型为long的变 量时，Java序列化机制会根据编译的class自动生成一个serialVersionUID作序列化版本比较用，这种情况下，只有同一次编译生成的 class才会生成相同的serialVersionUID 。
如果我们不希望通过编译来强制划分软件版本，即实现序列化接口的实体能够兼容先前版本，未作更改的类，就需要显式地定义一个名为serialVersionUID，类型为long的变量，不修改这个变量值的序列化实体都可以相互进行串行化和反串行化。
你可以很轻松的在google上查询到详细的解释。