开发者

How can I make a named query with IN actually work?

class X {
Y y; // manyToOne
}
class Y {
Long id;
}

@NamedQuery(name = "someName", query = "from X where y.id in :ids")

I have public, table, entity and all other things on the Entities but I didn't wrote them here.

TypedQuery<X> query = getEntityManager().createNamedQuery("someName", X.class);
query.setParameter("ids", someListOfLongs); // HERE I GET THE ERROR
queryFinal.getResultList();

Parameter value [[Ljava.lang.Object;@90d0bf] was not matching type [java.lang.Long]

I tried with or without (), I changed 开发者_JAVA技巧the version of Hibernate-Core to 3.6.4 (from JBoss 6.0.0.Final), otherwise if I wrote in :ids without () I'd got an error.

Please Help.

The IN always worked, the problem was that List<Long> wasn't actually List<Long> was List<Object[]>. Thanks


I also use JBoss AS 6 and this exact construct, but it just works.

This is an example of a query:

<named-query name="Item.getByItemIDs">
    <query>
        SELECT
            i
        FROM
            Item i
        WHERE
            i.ID in (:itemsIDs)
    </query>
</named-query>

And of a class using it:

@Override
public List<Item> getByItemIDs(List<Long> itemIDs) {
    return entityManager.createNamedQuery("Item.getByItemIDs", Item.class)
                        .setParameter("itemIDs", itemIDs)
                        .getResultList();
}

As your exception indicated [[Ljava.lang.Object;@90d0bf] (which is an Object[]), maybe you should try List<Long> as in my example?

(p.s. You can use the fluid API of JPA to make your code a little less verbose)


The 'IN' clause in hibernate is very, very tricky, and as far as I remember doesn't play well with named parameters, but it does with positional parameters.

Try changing the query to this

@NamedQuery(name = "someName", query = "select x from X x where x.y.id in (?)")

and the code that uses it to

TypedQuery<X> query = getEntityManager().createNamedQuery("someName", X.class);
query.setParameter(1, someListOfLongs); // I can't remember if the position is 0 or 1 based.
queryFinal.getResultList();

And I should add that if the list is emtpy, you'll get an exception. In our code, we detect this situation and replace the empty list with a list that contains a '-1' and since all ids are positive we don't get any results back.

0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜