5

我有这样的课，

class SampleClass implements Serializable {
    String name;
    Serializable fieldName;
}

另一个类，例如，

class AnotherClass implements Serializable {
    SampleClass sampleClass;
}

这两个类都有自己的 getter 和 setter 方法。

在主类中，我sampleClass从 getter 函数中获取变量，并尝试使用这些sampleClass对象。但是当我使用它时，我遇到了类似的错误，could not deserialize.

如何访问成员SampleClass，或者我们应该有类型的字段成员Serializable？

谢谢。

编辑：我正在使用休眠，它使用 aemploye 和 aaddress 表之间的多对一关系。

我为上述两个表创建了 Hibernate 配置文件和 net beans 中的逆向工程文件。

然后我生成了 POJO 类。

类和xml是：

Aaddress.hbm.xml

<hibernate-mapping>
<class name="hibernatetutor.tablebeans.Aaddress" table="aaddress" schema="public">
    <id name="sno" type="int">
        <column name="sno" />
        <generator class="assigned" />
    </id>
    <property name="street" type="serializable">
        <column name="street" />
    </property>
    <set name="aemployes" inverse="true">
        <key>
            <column name="address" />
        </key>
        <one-to-many class="hibernatetutor.tablebeans.Aemploye" />
    </set>
</class>

Aemploye.hbm.xml

<hibernate-mapping>
<class name="hibernatetutor.tablebeans.Aemploye" table="aemploye" schema="public">
    <id name="id" type="int">
        <column name="id" />
        <generator class="assigned" />
    </id>
    <many-to-one name="aaddress" class="hibernatetutor.tablebeans.Aaddress" fetch="select">
        <column name="address" />
    </many-to-one>
    <property name="name" type="string">
        <column name="name" />
    </property>
</class>

Aaddress.java

public class Aaddress implements java.io.Serializable {

    private int sno;
    private Serializable street;
    private Set aemployes = new HashSet(0);

    public int getSno() {
        return this.sno;
    }

    public void setSno(int sno) {
        this.sno = sno;
    }

    public Serializable getStreet() {
        return this.street;
    }

    public void setStreet(Serializable street) {
        this.street = street;
    }

    public Set getAemployes() {
        return this.aemployes;
    }

    public void setAemployes(Set aemployes) {
        this.aemployes = aemployes;
    }
}

Aemploye.java

public class Aemploye implements java.io.Serializable {

    private int id;
    private Aaddress aaddress;
    private String name;

    public int getId() {
        return this.id;
    }

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

    public Aaddress getAaddress() {
        return this.aaddress;
    }

    public void setAaddress(Aaddress aaddress) {
        this.aaddress = aaddress;
    }

    public String getName() {
        return this.name;
    }

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

Main.java

private void getData() {
    Session session = HibernateUtils.getInstance().openSession();
    Query query = session.createQuery("from Aemploye where id=:id");
    query.setParameter("id", 1);
    Aemploye a = (Aemploye) query.uniqueResult();
    Aaddress a1 = a.getAaddress();
    System.out.println(a1.getStreet());
}

错误是：

org.hibernate.type.SerializationException: could not deserialize
    at org.hibernate.util.SerializationHelper.deserialize(SerializationHelper.java:217)
    at org.hibernate.util.SerializationHelper.deserialize(SerializationHelper.java:240)
    at org.hibernate.type.SerializableType.fromBytes(SerializableType.java:82)
    at org.hibernate.type.SerializableType.get(SerializableType.java:39)
    at org.hibernate.type.NullableType.nullSafeGet(NullableType.java:163)
    at org.hibernate.type.NullableType.nullSafeGet(NullableType.java:154)
    at org.hibernate.type.AbstractType.hydrate(AbstractType.java:81)
    at org.hibernate.persister.entity.AbstractEntityPersister.hydrate(AbstractEntityPersister.java:2096)
    at org.hibernate.loader.Loader.loadFromResultSet(Loader.java:1380)
    at org.hibernate.loader.Loader.instanceNotYetLoaded(Loader.java:1308)
    at org.hibernate.loader.Loader.getRow(Loader.java:1206)
    at org.hibernate.loader.Loader.getRowFromResultSet(Loader.java:580)
    at org.hibernate.loader.Loader.doQuery(Loader.java:701)
    at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:236)
    at org.hibernate.loader.Loader.loadEntity(Loader.java:1860)
    at org.hibernate.loader.entity.AbstractEntityLoader.load(AbstractEntityLoader.java:48)
    at org.hibernate.loader.entity.AbstractEntityLoader.load(AbstractEntityLoader.java:42)
    at org.hibernate.persister.entity.AbstractEntityPersister.load(AbstractEntityPersister.java:3044)
    at org.hibernate.event.def.DefaultLoadEventListener.loadFromDatasource(DefaultLoadEventListener.java:395)
    at org.hibernate.event.def.DefaultLoadEventListener.doLoad(DefaultLoadEventListener.java:375)
    at org.hibernate.event.def.DefaultLoadEventListener.load(DefaultLoadEventListener.java:139)
    at org.hibernate.event.def.DefaultLoadEventListener.onLoad(DefaultLoadEventListener.java:98)
    at org.hibernate.impl.SessionImpl.fireLoad(SessionImpl.java:878)
    at org.hibernate.impl.SessionImpl.immediateLoad(SessionImpl.java:836)
    at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:66)
    at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:111)
    at org.hibernate.proxy.pojo.cglib.CGLIBLazyInitializer.invoke(CGLIBLazyInitializer.java:150)
    at hibernatetutor.tablebeans.Aaddress$$EnhancerByCGLIB$$44bec229.getStreet(<generated>)
    at hibernatetutor.Main.getData(Main.java:33)
    at hibernatetutor.Main.main(Main.java:24)
Caused by: java.io.StreamCorruptedException: invalid stream header
    at java.io.ObjectInputStream.readStreamHeader(ObjectInputStream.java:753)
    at java.io.ObjectInputStream.<init>(ObjectInputStream.java:268)
    at org.hibernate.util.SerializationHelper$CustomObjectInputStream.<init>(SerializationHelper.java:252)
    at org.hibernate.util.SerializationHelper.deserialize(SerializationHelper.java:209)
    ... 29 more
4

3 回答 3

1

根据问题和评论部分的一些信息，我相信您的麻烦是由以下原因引起的：

出于某种原因，您将 street 属性选择为可序列化类型。在您的表中，该列已定义为类型TEXT。Hibernate 可能设法将序列化的数据保存到列中，但数据库可能无法使它们保持不变。因此，在检索时，现在出现乱码的序列化无法反序列化。

正如 Petr Pudlák 所指出的，解决方案是让您的映射正确。如果您选择合适的二进制类型，例如BYTEA，那么您将能够原封不动地存储二进制数据。然后检索应该工作。

恕我直言，这不是正确的解决方案，首先是在您的 java 代码中选择合适的数据类型。将街道类型设为 Serializable 会让任何查看您的代码的人感到困惑。String可能更有意义，并且也非常适合 column type TEXT。

于 2012-08-09T12:20:16.590 回答
0

除非您有一些非常不寻常的框架来执行此操作，否则使用 getter 不涉及序列化。

我建议您查看确切的堆栈跟踪（并将其发布在问题中）并查看异常实际发生的位置。

于 2012-08-09T07:57:33.690 回答
0

我试过你的课，它对我有用：

import java.io.*;

class SampleClass implements Serializable {
      String name;
      Serializable fieldName;
}

class AnotherClass implements Serializable {
      SampleClass sampleClass;
}

public class Ser {
    public static void main(String argv[])
        throws Exception
    {
        SampleClass s = new SampleClass();
        s.name = "name";
        s.fieldName = "fieldName";

        AnotherClass a = new AnotherClass();
        a.sampleClass = s;

        // serialize the classes to a byte array
        ByteArrayOutputStream os = new ByteArrayOutputStream();
        ObjectOutputStream oos = new ObjectOutputStream(os);
        oos.writeObject(a);
        oos.close();

        // deserialize the classes from the byte array
        ObjectInputStream is
            = new ObjectInputStream(
                    new ByteArrayInputStream( os.toByteArray() ));
        a = (AnotherClass)is.readObject();
        is.close();

        // print something
        System.out.println(a.sampleClass.name);
    }
}

您可以发布导致问题的确切代码吗？

于 2012-08-09T07:58:25.040 回答