Spring CrudRepository throws org.hibernate.LazyInitializationException

Question

i got a problem by using CrudRepository. Example: i have two entities, entity A has a collection of entity B.

class A {
  int id;
  int name;
  @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)
  Set<B> bs;
  // getters and setters
}
class B {
  int id;
  int name;
  @ManyToOne(mappedBy="bs")
  A a;
  // getters and setters
}

then i got 2 repositories.

class ARepository extends CrudRepository<A, int>{}
class BRepository extends CrudRepository<B, int>{}

but when i got this, i got a org.hibernate.LazyInitializationException, how can i avoid this?

@Service
@Transactional(readOnly=true)
class ServiceImpl implements Service {
@Resource ARepository ar;

@Override
A a = ar.findOne(int id);
}

here is the applicationContext.xml:

<jpa:repositories base-package="com.myproject.repository" />

<context:component-scan base-package="com.myproject.*" />
<context:annotation-config />

<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager" >
    <property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
<bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
    <property name="dataSource" ref="dataSource" />
    <property name="persistenceUnitName" value="keep-apm" />
    <property name="jpaVendorAdapter">
        <bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
            <property name="showSql" value="true" />
            <property name="generateDdl" value="true" />
            <property name="database" value="POSTGRESQL"/>
            <property name="databasePlatform" value="org.hibernate.dialect.PostgreSQLDialect"/>
        </bean>
    </property>
</bean>
<bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
    <property name="username" value="root" />
    <property name="password" value="root" />
    <property name="driverClassName" value="org.postgresql.Driver" />
    <property name="url" value="jdbc:postgresql://127.0.0.1:5432/db" />
</bean>
<bean id="sessionFactory" factory-bean="entityManagerFactory" factory-method="getSessionFactory">
</bean>
<tx:annotation-driven transaction-manager="transactionManager" />

here is the web.xml

<filter>
    <filter-name>openSessionInViewFilter</filter-name>
    <filter-class>org.springframework.orm.hibernate4.support.OpenSessionInViewFilter</filter-class>
    <init-param>
        <param-name>singleSession</param-name>
        <param-value>true</param-value>
    </init-param>

</filter>
<filter-mapping>
    <filter-name>openSessionInViewFilter</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>
<filter>
    <filter-name>struts2</filter-name>
    <filter-class>org.apache.struts2.dispatcher.ng.filter.StrutsPrepareAndExecuteFilter</filter-class>
</filter>
<filter-mapping>
    <filter-name>struts2</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>
<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<listener>
    <listener-class>com.myproject.util.LogLocator</listener-class>
</listener>

<welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
</welcome-file-list>

a.bs(the collection) would not be loaded, and always throw out a org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: no session or session was closed

Thank you in advance!!

Sean Patrick Floyd · Accepted Answer · 2012-07-02 12:52:27Z

2

You need to use start a transaction, preferably through Spring's Declarative Transaction Management.

In most cases:

define a TransactionManager in your XML
add this to your XML <tx:annotation-driven />
annotate your service method with @Transactional

You can find a sample setup in section Using @Transactional

answered Jul 2, 2012 at 12:52

Sean Patrick Floyd

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1 Comment

Jaiwo99 Over a year ago

Hi Sean, thanks for answering me, but i have already done, everything. i just added more details. please help me with this.

Jaiwo99 · Accepted Answer · 2012-07-10 09:55:29Z

0

Here is the solution:

i can not access the object on my web layer, i should access the lazy loading objects in service layer, which should be in a transaction.

answered Jul 10, 2012 at 9:55

Jaiwo99

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1 Comment

Lion Over a year ago

You must be able to access lazily loaded objects in a web view through the use of openSessionInViewFilter because it keeps the session open until the request completes in its entirely. Also, you should not be using singleSession=true. It causes problems (it however does not provide a first-level cache for the entire request).

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