所以我有这两节课:
样品
@Entity
@Table(name="sample")
public class Sample implements Serializable {
@Id
@GeneratedValue
@Column(name="sample_id")
private Long sample_id;
@Column(name="id")
private String id;
@Column(name="description")
private String description;
@ManyToOne
@JoinColumn(name="dna_study_id")
private DNA_Study study;
...Getters and setters ...
DNA_研究
@Entity
@Table(name = "dna_study")
public class DNA_Study implements Serializable {
@Id
@GeneratedValue
@Column(name="dna_study_id")
private Long id;
@Column(name="name")
private String name;
@Column(name="description")
private String description;
@Column(name="date")
private Date date;
@OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
@JoinColumn(name = "dna_study_id")
private List<Sample> samples;
我想从我的数据库中获取所有DNA_Study,并具有以下DAO:
@Repository
public interface DNA_StudyDAO extends CrudRepository<DNA_Study, Long>{ }
和这个RestController:
@RestController
public class AnalysisController {
ClassPathXmlApplicationContext context;
@CrossOrigin
@RequestMapping("/getanalysis")
public ArrayList<DNA_Study> getAnalysis() {
context = new ClassPathXmlApplicationContext("applicationContext.xml");
DNA_StudyDAO dao = context.getBean(DNA_StudyDAO.class);
return (ArrayList<DNA_Study>) dao.findAll();
}
当我调用它时,我得到“无法编写内容:无法延迟初始化角色的集合”,我试图修改我的DAO,因此方法findAll()更改为:
@Override
@Query("select d from DNA_Study d join fetch d.samples")
Iterable<DNA_Study> findAll();
使用此方法不会引发异常,但是调用方法会创建一个无限循环,因为创建DNA_Study意味着加载它的样本,每个样本加载它的DNA_Study等,因此会中断。所以我假设我需要添加一个OpenEntityManagerInViewFilter并撤消对findAll()的覆盖,尝试添加到我的SpringBootServletInitializer类中:
@Override
public void onStartup(ServletContext servletContext) throws ServletException
{
AnnotationConfigWebApplicationContext rootContext = new AnnotationConfigWebApplicationContext();
rootContext.register(Application.class);
rootContext.setServletContext(servletContext);
ServletRegistration.Dynamic dispatcher = servletContext.addServlet("dispatcher", new DispatcherServlet(rootContext));
dispatcher.setLoadOnStartup(1);
dispatcher.addMapping("/");
FilterRegistration.Dynamic filter = servletContext.addFilter("openEntityManagerInViewFilter", OpenEntityManagerInViewFilter.class);
filter.setInitParameter("singleSession", "true");
filter.addMappingForServletNames(null, true, "dispatcher");
servletContext.addListener(new ContextLoaderListener(rootContext));
}
但是当我打电话时,我仍然会“无法延迟初始化角色集合”
我应该如何正确添加OpenEntityManagerInViewFilter?
编辑
扩展SpringBootServletInitializer的类:
@Override
protected SpringApplicationBuilder configure(SpringApplicationBuilder application) {
return application.sources(GemDomusServerApplication.class);
}
@Override
public void onStartup(ServletContext servletContext) throws ServletException
{
AnnotationConfigWebApplicationContext rootContext = new AnnotationConfigWebApplicationContext();
rootContext.register(Application.class);
rootContext.setServletContext(servletContext);
ServletRegistration.Dynamic dispatcher = servletContext.addServlet("dispatcher", new DispatcherServlet(rootContext));
dispatcher.setLoadOnStartup(1);
dispatcher.addMapping("/");
FilterRegistration.Dynamic filter = servletContext.addFilter("openEntityManagerInViewFilter", OpenEntityManagerInViewFilter.class);
filter.setInitParameter("singleSession", "true");
filter.setInitParameter("entityManagerFactoryBeanName", "entityManagerFactory");
filter.setInitParameter("flushMode", "auto");
filter.addMappingForServletNames(null, true, "dispatcher");
servletContext.addListener(new ContextLoaderListener(rootContext));
servletContext.addListener(new RequestContextListener());
}
public static void main(String[] args) {
SpringApplication.run(GemDomusServerApplication.class, args);
}
涉及的applicationContext.xml bean
<bean id="entityManagerFactory"
class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="dataSource" ref="dataSource" />
<property name="persistenceUnitName" value="jpaData" />
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter" />
</property>
<property name="jpaProperties">
<props>
<prop key="hibernate.dialect">org.hibernate.dialect.PostgreSQLDialect</prop>
<prop key="hibernate.show_sql">true</prop>
<prop key="hibernate.format_sql">true</prop>
<prop key="hibernate.hbm2ddl.auto">update</prop>
</props>
</property>
</bean>
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
<bean id="dataSource"
class="org.springframework.jdbc.datasource.DriverManagerDataSource">
<property name="driverClassName">
<value>org.postgresql.Driver</value>
</property>
<property name="url">
<value>**</value>
</property>
<property name="username">
<value>**</value>
</property>
<property name="password">
<value>**</value>
</property>
</bean>
persistence.xml
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="1.0"
xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd">
<persistence-unit name="jpaData"/>
</persistence>
最佳答案
您的问题与这个Lazy Initialisation with OpenEntityManagerInViewFilter?非常相似
我的另一个选择是,在您的DAO或管理器中调用Hibernate.initialize(Object),用@Transactional注释您的DAO。