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java - hibernate 组织. hibernate .LazyInitializationException : failed to lazily initialize a collection of role:

转载 作者:塔克拉玛干 更新时间:2023-11-03 04:45:46 27 4
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我有下面提到的实体类,当我执行我的应用程序时,出现以下异常。其他一些类似的问题没有解决问题。

WARNING: StandardWrapperValve[jersey-serlvet]: PWC1406: Servlet.service()
for servlet jersey-serlvet threw exception
org.hibernate.LazyInitializationException: failed to lazily initialize
a collection of role: test.entity.Dept.empDeptno, no session
or session was closed
at org.hibernate.collection.internal.AbstractPersistentCollection.
throwLazyInitializationException(AbstractPersistentCollection.java:393)
at org.hibernate.collection.internal.AbstractPersistentCollection.
throwLazyInitializationExceptionIfNotConnected
(AbstractPersistentCollection.java:385)
at org.hibernate.collection.internal.AbstractPersistentCollection.
initialize(AbstractPersistentCollection.java:378)

我该如何解决这个问题?

Emp 实体

@Entity
@Table(name = "EMP", schema = "SCOTT"
)
@XmlRootElement
@NamedQueries({
@NamedQuery(name = "Emp.findAllEmployees", query = "select e from Emp e left
join fetch e.deptNo order by e.empno desc")
})
public class Emp implements java.io.Serializable {
@Id
@Column(name = "EMPNO", unique = true, nullable = false, precision = 4,
scale = 0)
private short empno;
@ManyToOne
@JoinColumn(name = "DEPTNO", referencedColumnName = "DEPTNO")
private Dept deptNo;

部门实体

@Entity
@Table(name = "DEPT", schema = "SCOTT"
)
@XmlRootElement
public class Dept implements java.io.Serializable {
@Id
@Column(name = "DEPTNO", unique = true, nullable = false, precision = 2,
scale = 0)
private short deptno;
@OneToMany(fetch=FetchType.LAZY,mappedBy = "deptNo")
private Set<Emp> empDeptno;

DAOImpl

@Override
public List<Emp> findAllEmployees() {
return getEntityManager().createNamedQuery("Emp.findAllEmployees",
Emp.class).getResultList();
}

Jersey RESTful 服务

 @Component
@Path("/employee")
public class EmployeeRestService {

@Autowired
EmployeeService employeeService;

@GET
@Produces({MediaType.APPLICATION_JSON})
public List<Emp> getEmployees() {
List<Emp> emp = new ArrayList<Emp>();
emp.addAll(getEmployeeService().findAllEmployees());
return emp;
}

Spring applicationContext.xml

<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/tx
http://www.springframework.org/schema/tx/spring-tx-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd"
>
<!-- Data Source Declaration -->
<bean id="DataSource" class="org.springframework.jndi.JndiObjectFactoryBean">
<property name="jndiName" value="jdbc/scottDS"/>
</bean>

<context:component-scan base-package="net.test" />
<bean class="org.springframework.dao.annotation.PersistenceExceptionTranslationPostProcessor"/>
<bean id="entityManagerFactory"
class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="dataSource" ref="DataSource" />
<property name="packagesToScan" value="net.test" />
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
<property name="showSql" value="false" />
<property name="generateDdl" value="false" />
<property name="databasePlatform" value="${jdbc.dialectClass}" />
</bean>
</property>
</bean>
<bean id="defaultLobHandler" class="org.springframework.jdbc.support.lob.DefaultLobHandler" />
<!-- Transaction Config -->
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
<tx:annotation-driven transaction-manager="transactionManager"/>
<context:annotation-config/>
<bean id="hibernateStatisticsMBean" class="org.hibernate.jmx.StatisticsService">
<property name="statisticsEnabled" value="true" />
<property name="sessionFactory" value="#{entityManagerFactory.sessionFactory}" />
</bean>
</beans>

最佳答案

我已经通过在 web.xml 中添加以下内容解决了这个问题

<filter>
<filter-name>OpenEntityManagerInViewFilter</filter-name>
<filter-class>org.springframework.orm.jpa.support.OpenEntityManagerInViewFilter</filter-class>
</filter>
<filter-mapping>
<filter-name>OpenEntityManagerInViewFilter</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>

礼貌 herehere

谢谢

关于java - hibernate 组织. hibernate .LazyInitializationException : failed to lazily initialize a collection of role:,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20162077/

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