I' m trying to translate to Hibernate this Query:
SELECT n.*, (select nl.sid from newsReaded nl where nl.newsfk=n.sid and nl.reader='john') as readed
FROM news n

I use a subquery in the select clause, then i have read "Hibernate in action" to look for subselects.

I read that HQL supports subqueries only in the where clause, but also i read that we can use Formulas in the property clause to put it. It would look like this:
<property
name="readed"
formula="( select clause )"
type="long"/>

But my problem is that i need to pass an external parameter. Instead of: nl.reader='john' i need to put something like nl.reader=v_name

I´m using hibernate version 3.0.
Thanks in advance for your valuable time.
Sorry for my English.
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you can access the columns available in your mapped class.

so writing
... nl.reader = name
will refer to column name of your reader.
Have a look at the examples in the reference. This is explained there!

Please rate if this helped.
Regards
Sebastian

thanks for your reply... very thanks,
but the problem is not to access to any column name... the problem is that:
I use in the query an external parameter that i retrieved in the user session.
Thanks again for your valuable time.

SELECT n.*, nl.sid
FROM news n LEFT JOIN newsReaded nl ON nl.newsfk=n.sid and nl.reader='john'

I think this will not work with a mapping but with a query.

You could create a query using a subselect by yourself, so you will receive a tuple of objects.
obj[0] your object, obj[1] your additional value

Look for
Scalar results
and
subqueries
in the reference.

Regards Sebastian

baliukas, thanks for your query. I've tested queries with outer joins (like as yours and others) but i retrieve an erroneus results :(
and thanks also LaLiLuna.
I will use several queries instead of the unique query :S
I've tested that and the performance is not bad.