CODEWITHSUNDEEP
javahibernatejpacriteria-api
BartoszCichecki
BartoszCichecki

Reputation: 2267

JPA & Criteria API - Select only specific columns

I would like to select only specific columns (ex. SELECT a FROM b). I have a generic DAO and what I came up with is:

public List<T> getAll(boolean idAndVersionOnly) {
    CriteriaBuilder builder = manager.getCriteriaBuilder();
    CriteriaQuery<T> criteria = builder.createQuery(entityClazz);
    Root<T> root = criteria.from(entityClazz);
    if (idAndVersionOnly) {
        criteria.select(root.get("ID").get("VERSION")); // HERE IS ERROR
    } else {
        criteria.select(root);
    }
    return manager.createQuery(criteria).getResultList();
}

And the error is: The method select(Selection<? extends T>) in the type CriteriaQuery<T> is not applicable for the arguments (Path<Object>). How should I change that? I want to get a type T object that has only ID and VERSION fields, and all others are null.

Type T extends AbstractEntity which has those 2 fields.

entityClazz is T.class.

Upvotes: 88

Views: 168064

Answers (4)

Abrar Ansari
Abrar Ansari

Reputation: 147

You can do something like this

Session session = app.factory.openSession();
CriteriaBuilder builder = session.getCriteriaBuilder();
CriteriaQuery query = builder.createQuery();
Root<Users> root = query.from(Users.class);
query.select(root.get("firstname"));
String name = session.createQuery(query).getSingleResult();

where you can change "firstname" with the name of the column you want.

Upvotes: -3

EduardoRamirez
EduardoRamirez

Reputation: 161

cq.select(cb.construct(entityClazz.class, root.get("ID"), root.get("VERSION")));  // HERE IS NO ERROR

https://wiki.eclipse.org/EclipseLink/UserGuide/JPA/Basic_JPA_Development/Querying/Criteria#Constructors

Upvotes: 16

perissf
perissf

Reputation: 16273

One of the JPA ways for getting only particular columns is to ask for a Tuple object.

In your case you would need to write something like this:

CriteriaQuery<Tuple> cq = builder.createTupleQuery();
// write the Root, Path elements as usual
Root<EntityClazz> root = cq.from(EntityClazz.class);
cq.multiselect(root.get(EntityClazz_.ID), root.get(EntityClazz_.VERSION));  //using metamodel
List<Tuple> tupleResult = em.createQuery(cq).getResultList();
for (Tuple t : tupleResult) {
    Long id = (Long) t.get(0);
    Long version = (Long) t.get(1);
}

Another approach is possible if you have a class representing the result, like T in your case. T doesn't need to be an Entity class. If T has a constructor like:

public T(Long id, Long version)

then you can use T directly in your CriteriaQuery constructor:

CriteriaQuery<T> cq = builder.createQuery(T.class);
// write the Root, Path elements as usual
Root<EntityClazz> root = cq.from(EntityClazz.class);
cq.multiselect(root.get(EntityClazz_.ID), root.get(EntityClazz_.VERSION));  //using metamodel
List<T> result = em.createQuery(cq).getResultList();

See this link for further reference.

Upvotes: 111

Nikola Yovchev
Nikola Yovchev

Reputation: 10206

First of all, I don't really see why you would want an object having only ID and Version, and all other props to be nulls. However, here is some code which will do that for you (which doesn't use JPA Em, but normal Hibernate. I assume you can find the equivalence in JPA or simply obtain the Hibernate Session obj from the em delegate Accessing Hibernate Session from EJB using EntityManager ):

List<T> results = session.createCriteria(entityClazz)
    .setProjection( Projections.projectionList()
        .add( Property.forName("ID") )
        .add( Property.forName("VERSION") )
    )
    .setResultTransformer(Transformers.aliasToBean(entityClazz); 
    .list();

This will return a list of Objects having their ID and Version set and all other props to null, as the aliasToBean transformer won't be able to find them. Again, I am uncertain I can think of a situation where I would want to do that.

Upvotes: 3

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