1. JPA: Invalid-Key Error when using the Criteria API stackoverflow.com
I always get the same error when using the criteria API:
2. Criteria, Example.create () and foreign keys... forum.hibernate.org
Maybe I'm using it wrong, but it seems that Example.create () only works for simple properties (Long/String/etc) and doesn't take into account foreign key relationships. My use case: Address a = new Address () ; a.setCity ("somecity") ; a.setState (new State ("AL")) ; ... = Example.create (a) ; The State entity is never included in the generated SQL... I can implement ...
3. [CRITERIA] Join without foreign key forum.hibernate.org