CODEWITHSUNDEEP
javahibernatejpacriteria-api
Gazeciarz
Gazeciarz

Reputation: 525

JPA Criteria multiselect with fetch

I have following model:

@Entity
@Table(name = "SAMPLE_TABLE")
@Audited
public class SampleModel implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "ID")
    private Long id;

    @Column(name = "NAME", nullable = false)
    @NotEmpty
    private String name;

    @Column(name = "SHORT_NAME", nullable = true)
    private String shortName;

    @ManyToOne(fetch = FetchType.LAZY, optional = true)
    @JoinColumn(name = "MENTOR_ID")
    private User mentor;

//other fields here

//omitted getters/setters

}

Now I would like to query only columns: id, name, shortName and mentor which referes to User entity (not complete entity, because it has many other properties and I would like to have best performance).

When I write query:

CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery<SampleModel> query = builder.createQuery(SampleModel.class);
Root<SampleModel> root = query.from(SampleModel.class);
query.select(root).distinct(true);
root.fetch(SampleModel_.mentor, JoinType.LEFT);

query.multiselect(root.get(SampleModel_.id), root.get(SampleModel_.name), root.get(SampleModel_.shortName), root.get(SampleModel_.mentor));
query.orderBy(builder.asc(root.get(SampleModel_.name)));
TypedQuery<SampleModel> allQuery = em.createQuery(query);
return allQuery.getResultList();

I have following exception:

Caused by: org.hibernate.QueryException: query specified join fetching, but the owner of the fetched association was not present in the select list [FromElement{explicit,not a collection join,fetch join,fetch non-lazy properties,classAlias=generatedAlias1,role=com.sample.SampleModel.model.SampleModel.mentor,tableName=USER_,tableAlias=user1_,origin=SampleModel SampleModel0_,columns={SampleModel0_.MENTOR_ID ,className=com.sample.credential.model.User}}]
    at org.hibernate.hql.internal.ast.tree.SelectClause.initializeExplicitSelectClause(SelectClause.java:214)
    at org.hibernate.hql.internal.ast.HqlSqlWalker.useSelectClause(HqlSqlWalker.java:991)
    at org.hibernate.hql.internal.ast.HqlSqlWalker.processQuery(HqlSqlWalker.java:759)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:675)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:311)
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:259)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:262)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:190)
    ... 138 more

Query before exception:

SELECT DISTINCT NEW com.sample.SampleModel.model.SampleModel(generatedAlias0.id, generatedAlias0.name, generatedAlias0.shortName, generatedAlias0.mentor)
FROM com.sample.SampleModel.model.SampleModel AS generatedAlias0
LEFT JOIN FETCH generatedAlias0.mentor AS generatedAlias1
ORDER BY generatedAlias0.name ASC

I know that I can replace fetch with join but then I will have N+1 problem. Also I do not have back reference from User to SampleModel and I do not want to have..

Upvotes: 10

Views: 22767

Answers (3)

ozeray
ozeray

Reputation: 2254

It's been a long time since the question was asked. But I wish some other guys would benefit from my solution:

The trick is to use subquery.

Let's assume you have Applicant in your Application entity (one-to-one):

@Entity
public class Application {     

   private long id;
   private Date date;

   @OneToOne(fetch = FetchType.LAZY)
   @JoinColumn(name = "some_id")
   private Applicant applicant;

   // Other fields

   public Application() {}

   public Application(long id, Date date, Applicant applicant) {
       // Setters
   }
}

//...............

CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Application> cbQuery = cb.createQuery(Application.class);

Root<Application> root = cbQuery.from(Application.class);

Subquery<Applicant> subquery = cbQuery.subquery(Applicant.class);
Root subRoot = subquery.from(Applicant.class);

subquery.select(subRoot).where(cb.equal(root.get("applicant"), subRoot));
cbQuery.multiselect(root.get("id"), root.get("date"), subquery.getSelection());

This code will generate a select statement for Application, and select statements for Applicant per each Application.

Note that you have to define an appropriate constructor corresponding to your multiselect.

Upvotes: 4

Pierre C
Pierre C

Reputation: 3470

I got the same problem using EclipseLink as the JPA provider : I just wanted to return the id of a mapped entity («User» in Gazeciarz's example).

This can be achieved quite simply by replacing (in the query.multiselect clause)

root.get(SampleModel_.mentor)

with something like

root.get(SampleModel_.mentor).get(User_.id)

Then, instead of returning all the fields of User, the request will only return the its id.

I also used a tuple query but, in my case, it was because my query was returning fileds from more than one entity.

Upvotes: 0

maxenglander
maxenglander

Reputation: 4041

I ran into this same issue, and found that I was able to work around it by using:

CriteriaQuery<Tuple> crit = builder.createTupleQuery();

instead of

CriteriaQuery<X> crit = builder.createQuery(X.class);

A little extra work has to be done to produce the end result, e.g. in your case:

return allQuery.getResultList().stream()
    map(tuple -> {
        return new SampleModel(tuple.get(0, ...), ...));
    })
    .collect(toList());

Upvotes: 8

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