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Does Hibernate's Criteria API still not support nested relations

I'd like to use Hibernate's Criteria API for precisely what everybody says is probably its most likely use case, applying complex search criteria. Problem is, the table that I want to query against is not composed entirely of primitive values, but partially from other objects, and I need to query against those object's id's.

I found this article from 2 years ago that suggests it's not possible. Here's how I tried it to no avail, there are other aspect of Hibernate where I know of where this sort of dot notation is supported within string literals to indicate object nesting.

   if (!lookupBean.getCompanyInput().equals("")) {
       criteria.add(Restrictions.like("company.company", lookupBean.getCompanyInput() + "%"));
   }

EDIT:

Here's my correctly factored code for accomplishing what I was trying above, using the suggestion from the first answer below; note that I am even using an additional createCriteria call to order on an attribute in yet another associated object/table:

if (!lookupBean.getCompanyValue().equals("")) {
    criteria.createCriteria("company").add(
           Restrictions.like("company", lookupBean.getCompanyValue() + "%"));
}

List<TrailerDetail> tdList = 
        criteria.createCriteria("location").addOrder(Order.asc("location")).list();
like image 800
Dexygen Avatar asked Aug 14 '09 18:08

Dexygen


1 Answers

Not entirely sure I follow your example, but it's certainly possible to specify filter conditions on an associated entity, simply by nesting Criteria objects to form a tree. For example, if I have an entity called Order with a many-to-one relationship to a User entity, I can find all orders for a user named Fred with a query like this:

List<Order> orders = session.createCriteria(Order.class)
    .createCriteria("user")
        .add(eq("name", "fred"))
    .list();

If you're talking about an entity that has a relationship to itself, that should work as well. You can also replace "name" with "id" if you need to filter on the ID of an associated object.

like image 141
Rob H Avatar answered Sep 23 '22 15:09

Rob H