[hibernate-issues] [Hibernate-JIRA] Commented: (HHH-5735) Usage Path.type() in expressions throws java.lang.IllegalArgumentException

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 Fix For: 3.6.next, 4.0.0.Alpha1 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));