当数据正确显示时，为什么在此Spring MVC Web应用程序中出现Hibernate LazyInitializationException？

Question

我正在尝试使用Spring MVC创建一个Web应用程序，并将Hibernate作为其ORM层。但是，由于我对这两个框架的经验不足，所以我很努力。

以下代码将正确显示我要查找的所有记录，但仍将堆栈跟踪信息放入日志中。我在查找有关集成Hibernate和SpringMVC的详尽文档时遇到了麻烦（我查看了springsource.org并阅读了Internet上的各种文章）。谁能指出我在这里可能做错了什么？

请注意，为此，我花了一些时间试图在互联网上找到答案，包括查看此 SO问题。不幸的是，这没有帮助。

我还应该注意，该应用程序的ORM部分已在独立的Java应用程序中使用和测试，没有问题。因此，我相信Spring
MVC和Hibernate的集成是造成此问题的原因。

这是带有著名的延迟初始化问题的堆栈跟踪（被截断）；

2009-03-10 12:14:50,353 [http-8084-6] ERROR org.hibernate.LazyInitializationException.<init>(LazyInitializationException.java:19) - could not initialize proxy - no Session
org.hibernate.LazyInitializationException: could not initialize proxy - no Session
    at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:57)
    at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:111)
    at org.hibernate.proxy.pojo.cglib.CGLIBLazyInitializer.invoke(CGLIBLazyInitializer.java:150)
    at com.generic.orm.generated.SearchRule$$EnhancerByCGLIB$$92abaed6.toString(<generated>)
    at java.lang.String.valueOf(String.java:2827)
    at java.lang.StringBuffer.append(StringBuffer.java:219)
    at org.apache.commons.lang.builder.ToStringStyle.appendDetail(ToStringStyle.java:578)
    at org.apache.commons.lang.builder.ToStringStyle.appendInternal(ToStringStyle.java:542)
    at org.apache.commons.lang.builder.ToStringStyle.append(ToStringStyle.java:428)
    at org.apache.commons.lang.builder.ToStringBuilder.append(ToStringBuilder.java:840)
    at org.apache.commons.lang.builder.ReflectionToStringBuilder.appendFieldsIn(ReflectionToStringBuilder.java:606)
.....

这是我的网页控制器的代码；

private List<Report> getReports() {
    Session session = HibernateUtil.getSessionFactory().getCurrentSession();
    session.beginTransaction();

    List<Report> reports = session.createCriteria(Report.class).list();
    Hibernate.initialize(reports);

    session.getTransaction().commit();
    return reports;
}

使用此显示html在网页上使用的内容；

<table border="1">
    <c:forEach items="${model.reports}" var="report">
        <tr>
            <td><c:out value="${report.id}"/></td>
            <td><c:out value="${report.username}"/></td>
            <td><c:out value="${report.thresholdMet}"/></td>
            <td><c:out value="${report.results}"/></td>
            <td><c:out value="${report.searchRule.name}"/></td>
            <td><c:out value="${report.uuid}"/></td>
        </tr>
    </c:forEach>
</table>

注意：我添加了report.searchRule.name来测试是否可以到达report对象内的对象。它显示正常。

在我的applicationContext.xml中；

<bean id="sessionFactory" class="org.springframework.orm.hibernate3.LocalSessionFactoryBean">
    <property name="configLocation">
        <value>classpath:hibernate.cfg.xml</value>
    </property>
    <property name="hibernateProperties">
        <props>
            <prop key="hibernate.dialect">${hibernate.dialect}</prop>
        </props>
    </property>
</bean>

这是ORM映射，以防万一；

hibernate.cfg.xml（根据要求）

<hibernate-configuration>
  <session-factory>
    <property name="hibernate.connection.driver_class">com.microsoft.sqlserver.jdbc.SQLServerDriver</property>
    <property name="hibernate.connection.url">jdbc:sqlserver://<removed></property>
    <property name="hibernate.connection.username"><removed></property>
    <property name="hibernate.connection.password"><removed></property>
    <property name="hibernate.current_session_context_class">thread</property>
    <property name="hibernate.show_sql">false</property>
    <mapping resource="com/generic/orm/generated/Report.hbm.xml"/>
    <mapping resource="com/generic/orm/generated/FieldRule.hbm.xml"/>
    <mapping resource="com/generic/orm/generated/Reconciliation.hbm.xml"/>
    <mapping resource="com/generic/orm/generated/SearchRule.hbm.xml"/>
    <mapping resource="com/generic/orm/generated/IndexTemplate.hbm.xml"/>
    <mapping resource="com/generic/orm/generated/Field.hbm.xml"/>
    <mapping resource="com/generic/orm/generated/ErrorCode.hbm.xml"/>
  </session-factory>
</hibernate-configuration>

从report.hbm.xml

<hibernate-mapping>
    <class name="com.generic.orm.generated.Report" table="Report" schema="dbo" catalog="CoolRecon">
        <id name="id" type="int">
            <column name="ID" />
            <generator class="native" />
        </id>
        <timestamp name="timeStamp" column="TimeStamp" />
        <many-to-one name="searchRule" class="com.generic.orm.generated.SearchRule" fetch="select">
            <column name="SearchRuleName" length="50" not-null="true" />
        </many-to-one>
        <many-to-one name="errorCode" class="com.generic.orm.generated.ErrorCode" fetch="select">
            <column name="ErrorCodeId" />
        </many-to-one>
        <many-to-one name="reconciliation" class="com.generic.orm.generated.Reconciliation" fetch="select">
            <column name="ReconciliationName" length="100" />
        </many-to-one>
        <property name="username" type="string">
            <column name="Username" length="50" />
        </property>
        <property name="supersheetDate" type="timestamp">
            <column name="SupersheetDate" length="23" not-null="true" />
        </property>
        <property name="milliSecondsTaken" type="long">
            <column name="MilliSecondsTaken" not-null="true" />
        </property>
        <property name="thresholdMet" type="boolean">
            <column name="ThresholdMet" not-null="true" />
        </property>
        <property name="results" type="int">
            <column name="Results" not-null="true" />
        </property>
        <property name="exception" type="string">
            <column name="Exception" length="750" />
        </property>
        <property name="uuid" type="string">
            <column name="UUID" length="36" not-null="true" />
        </property>
    </class>
</hibernate-mapping>

Answer 1

我只是在猜测，但是从堆栈跟踪来看，似乎在SearchRule上调用了toString。SearchRule是否有任何未加载的子对象？如果SearchRule.toString试图获取未初始化的子对象的值，则可能导致LazyInitializationException。