[hibernate-issues] [Hibernate-JIRA] Created: (HHH-5675) Using compound primary key throws SQLException

[Ralf Pöhlmann (JIRA)]

Using compound primary key throws SQLException
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                 Key: HHH-5675
                 URL: http://opensource.atlassian.com/projects/hibernate/browse/HHH-5675
             Project: Hibernate Core
          Issue Type: Bug
          Components: core
    Affects Versions: 3.6.0, 3.6.0.CR2, 3.6.0.CR1, 3.5.6, 3.6.0.Beta4, 3.6.0.Beta3, 3.5.5, 3.6.0.Beta2, 3.6.0.Beta1, 3.5.4, 3.5.3, 3.5.2, 3.5.1, 3.5.0-Final
         Environment: Hibernate 3.6.0, MySql
            Reporter: Ralf Pöhlmann
            Priority: Critical


Using compound primary keys via @IdClass causes an exception when calling persist:

Caused by: javax.persistence.PersistenceException: org.hibernate.exception.GenericJDBCException: could not insert: [de.methodpark.stages.model.UserAssignment]
	at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1214)
	at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1147)
	at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1153)
	at org.hibernate.ejb.AbstractEntityManagerImpl.persist(AbstractEntityManagerImpl.java:678)
	at com.sun.enterprise.container.common.impl.EntityManagerWrapper.persist(EntityManagerWrapper.java:258)

Same code works when switching JPA provider to EclipseLink.

@Entity
@Table(name = "user_assignment")
@IdClass(value = UserAssignmentId.class)
public class UserAssignment implements Serializable {
	@Id
	private Long userId;

	@Id
	private Long elementId;

	@ManyToOne(fetch = FetchType.LAZY, optional = false)
	@JoinColumn(name = "element_id", referencedColumnName = "id")
	private Element element;

	@ManyToOne(fetch = FetchType.LAZY, optional = false)
	@JoinColumn(name = "user_id", referencedColumnName = "id")
	private User user;
}


public class UserAssignmentId implements Serializable {

	public UserAssignmentId() { }

	@Column(name = "user_id", insertable = false, updatable = false)
	public Long userId;

	@Column(name = "element_id", insertable = false, updatable = false)
	public Long elementId;
}

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