How to use hibernate criteria to return only one element of an object instead the entire object?

|▌冷眼眸甩不掉的悲伤 提交于 2019-11-30 06:05:59

I think you could do that with Projections, something like

Criteria.forClass(bob.class.getName())
        .add(Restrictions.gt("id", 10))
        .setProjection(Projections.property("id"))
        );

Similarly you can also:

Criteria criteria = session.createCriteria(bob.class);

criteria.add(Expression.gt("id", 10));

criteria.setProjection(Projections.property("id"));

criteria.addOrder(Order.asc("id"));

return criteria.list();

or setProjection(Projections.id())

SessionFactory sessionFactory;    
Criteria crit=sessionFactory.getCurrentSession().createCriteria(Model.class);
crit.setProjection(Projections.property("id"));
List result = crit.list();

This code code will give you list of ids in the model class like [1,2,3]. if you wants to get the array list like [{"id":1},{"id":2}] then use the following code

SessionFactory sessionFactory;    
Criteria crit=sessionFactory.getCurrentSession().createCriteria(Model.class); 
crit.setProjection(Projections.property("id").as("id")); 
List result = crit.setResultTransformer(Criteria.ALIAS_TO_ENTITY_MAP).list();

Another option (though a bit un hibernate-esque) is to use "raw" sql, like this:

List<Long> myList = session.createSQLQuery("select single_column from table_name")
          .addScalar("single_column", StandardBasicTypes.LONG).list();

You can do that like this

    bob bb=null;

    Criteria criteria = session.createCriteria(bob.class);  
    criteria.add(Restrictions.eq("id",id));

    bb = (bob) criteria.uniqueResult();

as Restrictions you can add your condition

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