Question

我有这样的课程OmQcActivity：

@Entity
@Cache(usage = CacheConcurrencyStrategy.READ_WRITE)
@Table(name="OM_QC_ACTIVITY")
public class OmQcActivity{
   @ManyToOne(fetch = FetchType.EAGER)
   @JoinColumn(name="STATUS_ID")
   private Codesc status;
}

codesc是另一个实体。

在我的代码中我写道：

OmQcActivity  myactivity = findQCActivityById(5);
Codesc status = myactivity.getCodesc();

@Transactional(readOnly = true, propagation = Propagation.SUPPORTS)
    public OmQcActivity findQCActivityById(Long id) {
        return session.load(persistentClass, id);
    }

然而，我得到了：

org.hibernate.LazyInitializationException: could not initialize proxy - no Session
    at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:86)
    at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:140)
    at org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer.invoke(JavassistLazyInitializer.java:190)
    at com.mycompany.model.OmQcActivity_$$_javassist_11.getStatus(OmQcActivity_$$_javassist_11.java)

如果获取类型是急切的，我怎么能得到Lazy异常？

Answer 1

load()不会立即加载实体，它会返回一个惰性代理，该代理在第一次方法调用时获取实际数据。在大多数情况下，您需要使用get()代替load()。

Answer 2

找到此link。如果Codesc实体中有CollectionOfElements注释，则会引发关于延迟初始化的异常。

当字段急切时，LazyInitializationException

2 个答案: