Hibernate Criteria for “in subselect”

2019-01-18 12:53发布

I'm trying to do something like this, but using Criteria instead of HQL:

select user from User where user in (
    select user from UserDomain where domain.id = "XXX"
)

User being an entity having a one-to-many relationship to the join table UserDomain. The point here is simply to find Users that are linked to a Domain having id = "XXX".

This seems like it should be very simple... but I'm having no luck so far turning up any useful docs.

3条回答
Ridiculous、
2楼-- · 2019-01-18 12:54

I finally found it. Turns out it wasn't so hard after all... once you know!

criteria = criteria.createCriteria(User.USER_DOMAINS).add(Restrictions.eq(UserDomain.DOMAIN, domain));

Yep, there is was, staring me right in the face in the Javadoc: http://www.dil.univ-mrs.fr/~massat/docs/hibernate-3.1/api/org/hibernate/Criteria.html

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Animai°情兽
3楼-- · 2019-01-18 13:00

The subquery is very useful in cases, that you need to search the User, having one-to-many UserDomains. In that case, the WHERE UserId IN (subquery) brings big advanatage: we are still working with a flat User table/entity... so we can do the proper paging.

Here is the documentation 15.8. Detached queries and subqueries

The draft could be: subquery:

DetachedCriteria userSubquery = DetachedCriteria.forClass(UserDomain.class, "ud")
    // Filter the Subquery
    .add(Restrictions.eq(UserDomain.DOMAIN, domain))
    // SELECT The User Id  
    .setProjection( Projections.property("ud.userId") );

And the main query:

Criteria query = session.createCriteria(User.class, "u")
    .add( Subqueries.propertyIn("u.id", userSubquery) );

Now we have a query, which could be used for paging

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太酷不给撩
4楼-- · 2019-01-18 13:05

In clauses usualy can be transformed into join. Tyr this:

Criteria c = session.createCriteria(User.class, "u");
c.createAlias("u.userDomain", "ud"); // inner join by default
c.add(Restrictions.le("ud.id", 1));
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