You are. A regular SelectQuery should work. You can use relationship
path to qualify on FK. E.g.
int id = ..;
ExpressionFactory.matchExp("artist", id);
Andrus
On Apr 26, 2010, at 2:32 PM, Joe Baldwin wrote:
> I asked this question previously but did not see a reply.
>
>
> I am trying to create a very fast custom fetch that is essentially a
> qualified relationship. My relationships are based on an integer
> "oid" field.
>
> Can I use the RelationshipQuery and qualify it? Or am I on the wrong
> track?
This archive was generated by hypermail 2.0.0 : Mon Apr 26 2010 - 04:44:15 EDT