Alex Rivera | Logout

EntityNotFoundException in Hibernate Many To One mapping however data exist

Asked 2012-11-24T06:50:15.397
60

I am getting javax.persistence.EntityNotFoundException error when I am trying to get User through Invoice object

invoice.getUser().getId()

Error is as follows

javax.persistence.EntityNotFoundException: Unable to find com.indianretailshop.domain.User with id 5
    at org.hibernate.ejb.Ejb3Configuration$Ejb3EntityNotFoundDelegate.handleEntityNotFound(Ejb3Configuration.java:137)
    at org.hibernate.proxy.AbstractLazyInitializer.checkTargetState(AbstractLazyInitializer.java:189)
    at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:178)
    at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:215)

Entity classes are as follows(getters and setters are not included)

@Entity
@Table(name="users")
public class User implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    @Column(unique=true, nullable=false)
    private int id;

    .
        .
        .

    //bi-directional many-to-one association to Invoice
    @OneToMany(mappedBy="user")
    private List<Invoice> invoices;
}

@Entity
@Table(name="invoice")
public class Invoice implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    @Column(unique=true, nullable=false)
    private int id;
        .
        .
        .

    //bi-directional many-to-one association to User
    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="Users_id")
    private User user;
}
Edit
Report

1 Answer

38

The problem could be that the direct entity does not exist, but also it could be that the referenced entity from that entity, normally for a EAGER fetch type, or optional=false.

Try this:

     //bi-directional many-to-one association to User
     @ManyToOne(fetch=FetchType.LAZY)
     @JoinColumn(name="Users_id")
     private User user = new User();
answered 2012-11-26T04:45:36.010

Your Answer