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? [英] Why am I getting a Hibernate LazyInitializationException in this Spring MVC web application when the data displays correctly?
问题描述
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都缺乏经验，所以一直在苦苦挣扎.
I am attempting to create a web application using Spring MVC, with Hibernate as its ORM layer. However, due to my inexperience with both frameworks I'm struggling.
以下代码将正确显示我正在查找的所有记录，但仍将堆栈跟踪放入我的日志中.我很难找到有关集成 Hibernate 和 SpringMVC 的完整文档(我在 springsource.org 上查看并阅读了互联网上的各种文章).谁能指出我在这里做错了什么?
The following code will properly display all the records I am looking for but still throw a stack trace into my logs. I'm having trouble finding thorough documentation concerning integrating Hibernate and SpringMVC (I've looked on springsource.org and read various articles on the interweb). Could anyone point out what I could be doing wrong here?
请注意，我花了一些时间试图在互联网上寻找答案，包括查看 this SO 问题.不幸的是，这没有帮助.
Please note that I have spent a few trying to track down answers on the internet for this, including looking at this SO question. Which was unfortunately no help.
我还应该注意到，这个应用程序的 ORM 部分已经在一个独立的 Java 应用程序中使用和测试，没有问题.所以我相信 Spring MVC 和 Hibernate 的集成导致了这个问题.
I should also note that the ORM part of this application has been used and tested in a stand alone Java application without problems. So I believe the integration of Spring MVC and Hibernate is causing the issue.
这是著名的延迟初始化问题的堆栈跟踪(截断)；
Here is the stack trace (truncated) with the famous lazy initialization issue;
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；
Here is a code from my web page controller;
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；
Which is employed on the web page using this display html;
<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="${report.uuid}"/></td>
</tr>
</c:forEach>
</table>
注意:我添加了 report.searchRule.name 来测试我是否可以获取报告对象中的对象.显示正常.
Note: That I added report.searchRule.name to test if I could get at the objects within the report object. It displays fine.
在我的 applicationContext.xml;
And in my applicationContext.xml;
<bean id="sessionFactory" class="org.springframework.orm.hibernate3.LocalSessionFactoryBean">
<property name="configLocation">
<value>classpath:hibernate.cfg.xml</value>
</property>
<property name="hibernateProperties">
<props>
<prop key="hibernate.dialect">${hibernate.dialect}</prop>
</props>
</property>
</bean>
这里是 ORM 映射，以防万一；
Here are the ORM mappings, just in case;
hibernate.cfg.xml(根据要求)
The hibernate.cfg.xml (as requested)
<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>
来自report.hbm.xml
From 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>
推荐答案
我只是猜测，但从堆栈跟踪来看，似乎在 SearchRule 上调用了 toString.SearchRule 是否有任何可能尚未加载的子对象?如果 SearchRule.toString 试图获取可能导致 LazyInitializationException 的未初始化子对象的值.
I am just guessing but from the stack trace it seems that toString is being called on SearchRule. Does SearchRule have any child objects that may not have been loaded? If SearchRule.toString was trying to get the value for an uninitialised child object that could result in the LazyInitializationException.
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