Usage Path.type() in expressions throws java.lang.IllegalArgumentException --------------------------------------------------------------------------- Key: HHH-5735 URL: http://opensource.atlassian.com/projects/hibernate/browse/HHH-5735 Project: Hibernate Core Issue Type: Bug Components: query-criteria Affects Versions: 3.6.0, 3.6.1 Reporter: Jaroslaw Lewandowski Priority: Blocker Attachments: InheritanceTest.tgz It is not possible to filter entities using Path.type() in Criteria Query API. Using type() in expressions throws an exception: Unexpected call on EntityTypeExpression#render java.lang.IllegalArgumentException: Unexpected call on EntityTypeExpression#render at org.hibernate.ejb.criteria.expression.PathTypeExpression.render( at org.hibernate.ejb.criteria.expression.PathTypeExpression.render(PathTypeExpression.java:48PathTypeExpression.java:48) ) at org.hibernate.ejb.criteria.predicate.ComparisonPredicate.render(ComparisonPredicate.java:173) at org.hibernate.ejb.criteria.QueryStructure.render(QueryStructure.java:258) at org.hibernate.ejb.criteria.CriteriaQueryImpl.render(CriteriaQueryImpl.java:340) at org.hibernate.ejb.criteria.CriteriaQueryCompiler.compile(CriteriaQueryCompiler.java:223) at org.hibernate.ejb.AbstractEntityManagerImpl.createQuery(AbstractEntityManagerImpl.java:441) at foo.InheritanceTest.critQryTest(InheritanceTest.java:55) Test case is attached and it's similar to an example from JPA 2.0 specification - chapter 6.5.7 Example 2: CriteriaQuery<Employee> q = cb.createQuery(Employee.class); Root<Employee> emp = q.from(Employee.class); q.select(emp) .where(cb.notEqual(emp.type(), Exempt.class)); -- This message is automatically generated by JIRA. - If you think it was sent incorrectly contact one of the administrators: http://opensource.atlassian.com/projects/hibernate/secure/Administrators.... - For more information on JIRA, see: http://www.atlassian.com/software/jira