JPA 2 + Criteria API + get oldest element in subquery

Question

I have two entities ( Dataset and Item ), where the first one contains a list of the latter one.

@Entity
class Dataset {
    @Id long id;
    List<Item> items;
}

@Entity
class Item {
    @Id long id;
    @Temporal Date date;
    String someCriteria;
}

I am now looking for any dataset , which references an item with someCriteria = 'x' as oldest item in the dataset's list (it's date is before that of all others referenced by this dataset).

This should work with the Criteria API of JPA-2.0, but an answer as SQL-query would be fine, too.

This is how it looks like right now:

Subquery<Item> _subquery = p_criteria.subquery(Item.class);
Root<Item> _itemRoot = _subquery.from(Item.class);
_subquery.select(_itemRoot);

// correlate
_subquery.correlate(_datsetRoot);

// type check
Predicate _typeP = p_builder.equal(_itemRoot.get(Item_.someCriteria), "x");

<additional predicates>
...

// reference check
_subquery.where(_typeP);

return p_builder.exists(_subquery);

I though about creating a subquery, order it by date and check the first one. Unfortunately JPA-2.0 does not allow ordering in subqueries (or joins) (as far as I can see). My second idea was getting MAX(date) , which should be supported by most of the databases. Same goes here, support for MAX, ABS, ... is only available for numbers.

EDIT: Using a subquery, which is sorted, it would look like this (MSSQL):

SELECT d.id
FROM Dataset d
WHERE EXISTS (
 SELECT TOP 1 *
 FROM Item i
 WHERE i.dataset_id = d.FALL_ID
    AND i.someCriteria = 'x'
 ORDER BY i.date
)
ORDER BY d.FALL_ID

(How) could I do this in JPA-2.0?

Answer 1

In Java 1.6 there are the methods CriteriaBuilder.least(expression) (instead of min() ) and CriteriaBuilder.lessThanOrEqualTo(expressionA, expressionB) (instead of le() ), which work for java.util.Date fields.

In Java 1.5 it seems that there is no way to achive it without additional fields in the DB.