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

[Jaroslaw Lewandowski (JIRA)]

Usage Path.type()  in expressions throws java.lang.IllegalArgumentException
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                 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));



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