JPA Criteria multiselect with fetch

Question

我有以下型号：

@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

}

现在我只想查询列：id、name、shortName 和 mentor 引用 User 实体（不是完整实体，因为它有很多其他属性，我希望有最好的性能）。

当我写查询时：

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();

我有以下异常：

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

异常前查询：

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

我知道我可以用 join 替换 fetch，但我会遇到 N+1 问题。此外，我没有从 User 到 SampleModel 的反向引用，我不想拥有..

Answer 1

我遇到了同样的问题，发现我可以使用以下方法解决它：

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

而不是

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

需要做一些额外的工作才能产生最终结果，例如在你的情况下：

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

Answer 2

这个问题已经很久没有问过了。但我希望其他一些人能从我的解决方案中受益：

诀窍是使用子查询。

假设您的 Application 实体中有申请人（一对一）：

@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());

此代码将为应用程序生成一个选择语句，并为每个应用程序的申请人选择语句。

请注意，您必须定义与您的多选相对应的适当构造函数。

Answer 3

我在使用 EclipseLink 作为 JPA 提供程序时遇到了同样的问题：我只想返回映射实体的 id（Gazeciarz 示例中的 «User»）。

这可以通过替换（在 query.multiselect 子句中）非常简单地实现

root.get(SampleModel_.mentor)

类似的东西

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

那么，请求不会返回User的所有字段，而是只返回它的id。

我也使用了元组查询，但就我而言，这是因为我的查询返回来自多个实体的文件。