Hello, all!

I have entity А with two foreign keys (FK) from Y, X.
There is also entity B, also with FKs from Y,X.
Database scema can't be changed.
I need a connection from A to B based on a condition that X and Y are equial in both A and B. I mean, i need to have B in A where a.x=b.x and a.y=b.y

I tried to do that using formula element in mapping:
<class name="A" table="a">
<property name='x' class='X' />
<property name='y' class='Y' />
<one-to-one name="b" class="B" formula="from B b where b.x=x and b.y=y" />
</class>
and
<class name="A" table="a">
<property name='x' class='X' />
<property name='y' class='Y' />
<one-to-one name="b" class="B" >
<formula>id</formula>
<formula>(select b.id from B where b.x=x and b.y=y)</formula>
</one-to-one>
</class>

In both cases I got errors when tried to load A. What's wrong with my formulas? Any help will be appretiated.

Regards, Andrey Grishin.

try to use sql instead of hql..Use table names and field names in place of bean class and its properties..

try to use sql instead of hql..Use table names and field names in place of bean class and its properties..

try to use sql instead of hql..Use table names and field names in place of bean class and its properties..

try to use sql instead of hql..Use table names and field names in place of bean class and its properties..

The formula element/attribute uses plain SQL, not HQL, so your first example won't work.

one-to-one uses primary key (id) properties, and as you can't use that, you must use many-to-one unique="true".

Are X and Y mapped? If they're just legacy tables that aren't being mapped, then you can just use a straightforward many-to-one unique="true" property-ref="xid" or whatever (though you'll need to "merge" the xid and yid columns into one hibernate property using the properties element; see section 5.1.14 of the ref docs).

If X and Y are mapped, then you don't need to model the relationship between A and B. Use queries to get corresponding items:

Obviously that query assumes that all relationships are bidirectional, but equivalents can be found if any of the relationships are unidirectional.