我有以下模型实体:
@Entity
@Table(name="ABONADO_IMEI")
public class SubscriberImei {
private SubscriberImeiId id;
private Terminal terminal;
private String state;
private Date date;
@EmbeddedId
public SubscriberImeiId getId() {
return id;
}
public void setId(SubscriberImeiId id) {
this.id = id;
}
@ManyToOne(fetch=FetchType.LAZY)
@JoinColumn(name="ID_TERMINAL", nullable=false)
public Terminal getTerminal() {
return terminal;
}
public void setTerminal(Terminal terminal) {
this.terminal = terminal;
}
@Column(name="Estado")
public String getState() {
return state;
}
public void setState(String state) {
this.state = state;
}
@Column(name="Fecha")
public Date getDate() {
return date;
}
public void setDate(Date date) {
this.date = date;
}
}
@Entity()
public class Terminal {
private long id;
private String code;
private String brand;
private String model;
private String services;
private Set<SubscriberImei> subscriberImei;
public Terminal(){
}
@Id
@SequenceGenerator(name="SEQ_TERMINAL_ID", sequenceName="SEQ_TERMINAL_ID",allocationSize=1)
@GeneratedValue(strategy=GenerationType.SEQUENCE, generator = "SEQ_TERMINAL_ID")
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
@Column(name="Codigo")
public String getCode() {
return code;
}
public void setCode(String code) {
this.code = code;
}
@Column(name="Marca")
public String getBrand() {
return brand;
}
public void setBrand(String brand) {
this.brand = brand;
}
@Column(name="Modelo")
public String getModel() {
return model;
}
public void setModel(String model) {
this.model = model;
}
@Column(name="Servicios")
public String getServices() {
return services;
}
public void setServices(String services) {
this.services = services;
}
@OneToMany(fetch=FetchType.LAZY , mappedBy="terminal")
public Set<SubscriberImei> getSubscriberImei() {
return subscriberImei;
}
public void setSubscriberImei(Set<SubscriberImei> terminals) {
this.subscriberImei = terminals;
}
}
我正在使用以下JPARepository访问它:
public interface SubscriberImeiDao extends JpaRepository<SubscriberImei, SubscriberImeiId>{
@Query("select u from SubscriberImei u where u.id.snb = ?1 and u.state = ?2")
SubscriberImei findBySnbAndState(String snb, String state);
}
我需要返回findBySnbAndState的结果。这是我的代码:
@Transactional
public SubscriberImei imeiQuery(String countryId, String snb) {
SubscriberImei subs = null;
try{
this.validateRegion(countryId, snb);
subs = this.getSubscriberDao().findBySnbAndState(snb,"A");
subs.getTerminal().getBrand();
if( subs != null)
log.info("Subscriber found {}", subs);
else
log.info("Subcriber not found!");
} catch (Exception ex) {
log.error("Unknown Exception -- ", ex);
}
return subs;
}
我在subs.getTerminal()。getBrand()行中遇到了提到的异常。通过这一行,我强迫遵循惰性关系(如其他一些问题所建议)。我还在上下文中添加了以下行:
<tx:annotation-driven mode="aspectj"/>
将FetchType更改为Eager可以使其正常工作,但我不想这样做,因为不必始终遵循这种关系。我想念什么?我究竟做错了什么?
谢谢
这是完整的异常stacktrace
org.hibernate.LazyInitializationException: could not initialize proxy - no Session
at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:165)
at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:286)
at org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer.invoke(JavassistLazyInitializer.java:185)
at ar.com.redmondsoftware.imeitracking.model.Terminal_$$_jvste48_0.getBrand(Terminal_$$_jvste48_0.java)
at ar.com.redmondsoftware.imeitrackingbusinesslogic.service.impl.ImeiQueryImpl.imeiQuery(ImeiQueryImpl.java:24)
at ar.com.redmondsoftware.imeitracking.QueryImei.testQueryImei(QueryImei.java:34)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:606)
at org.junit.runners.model.FrameworkMethod$1.runReflectiveCall(FrameworkMethod.java:50)
at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:12)
at org.junit.runners.model.FrameworkMethod.invokeExplosively(FrameworkMethod.java:47)
at org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:17)
at org.junit.internal.runners.statements.RunBefores.evaluate(RunBefores.java:26)
at org.junit.runners.ParentRunner.runLeaf(ParentRunner.java:325)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:78)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:57)
at org.junit.runners.ParentRunner$3.run(ParentRunner.java:290)
at org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:71)
at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:288)
at org.junit.runners.ParentRunner.access$000(ParentRunner.java:58)
at org.junit.runners.ParentRunner$2.evaluate(ParentRunner.java:268)
at org.junit.runners.ParentRunner.run(ParentRunner.java:363)
at org.eclipse.jdt.internal.junit4.runner.JUnit4TestReference.run(JUnit4TestReference.java:50)
at org.eclipse.jdt.internal.junit.runner.TestExecution.run(TestExecution.java:38)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:467)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:683)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.run(RemoteTestRunner.java:390)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.main(RemoteTestRunner.java:197)
最佳答案
如果您不希望它一直渴望获取,可以进行其他查询以在需要时进行渴望获取:
@Query("SELECT u FROM SubscriberImei u LEFT JOIN FETCH u.terminal WHERE u.id.snb = ?1 and u.state = ?2")
SubscriberImei findBySnbAndStateWithTerminal(String snb, String state);
几乎总是首选此查询,因为它将用一个查询而不是两个查询加载数据。
关于java - JPA存储库org.hibernate.LazyInitializationException:无法初始化代理-没有 session ,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/33486789/