1. JPA one-to-many association to an entity with @Inheritance(strategy=InheritanceType.JOINED) stackoverflow.com
I'm trying to map an association one-to-many to an entity that is mappes with
2. JPA @Inheritance(strategy = InheritanceType.JOINED): different discriminator value for same class stackoverflow.com
Is it possibile to map two differents discriminator values to the same concrete java class? Something like this:
Another question: is it possible to have ...
3. JPA @Inheritance(strategy = InheritanceType.JOINED): Missing class for indicator field value stackoverflow.com
Is it possible to have a default implementation when no class is provided for an indicator field value for not raising this exception?
Kind regards Massimo ...
4. JPA - InheritanceType.Joined generates wrong tables stackoverflow.com
I'm trying to represent in data base a editable table, with multiple field types. My approach looks like this:
5. Hibernate 4: persisting InheritanceType.JOINED discriminator column values stackoverflow.com
I have a simple JOINED hierarchy of documents:
6. Please explain below queries in case of JPA @Inheritance(strategy=InheritanceType.JOINED) coderanch.com
I have two entities - Product.java and Book.java. Book is extending Product and i have defiend inheritence as @Inheritance(strategy=InheritanceType.JOINED). As per my understanding, i will create two tables :- Product and Book. When i am trying to persist Book in the database using em.persist, it is inserting one record in Book as well as Product table. My question is that 1. ...
7. @Inheritance(strategy=InheritanceType.JOINED) and .class forum.hibernate.org
Hi Everyone, I am trying to use the @Inheritance(strategy=InheritanceType.JOINED) annotation with this query: @NamedQuery(name="RbacRole.getPermissionsByType", query="select p " + "from RbacRole as role " + "join role.permissions as p " + "where role.pk = :pk and p.class = :clazz") when I execute the query I get the following exception: java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Integer on the class property. When I ...