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我试图转换这个 JPQL 查询；

SELECT s FROM QuestionSet s JOIN s.questions q WHERE q.appointedRepetition.date  < :tomorrow

与其标准 api 等效，这是我到目前为止所拥有的：

    DateTime tomorrow = DateTime.now().plusDays(1).withTime(0,0,0,0);

    CriteriaBuilder criteriaBuilder = JPA.em().getCriteriaBuilder();
    CriteriaQuery<QuestionSet> query = criteriaBuilder.createQuery(QuestionSet.class);
    Root<QuestionSet> root = query.from(QuestionSet.class);
    Join<QuestionSet, Question> questionJoin = root.join("questions");
    Predicate ownerCondition = criteriaBuilder.equal(root.get("owner"), owner);
    Predicate dateCondition = criteriaBuilder.lessThan(questionJoin.<DateTime>get("appointedRepetition.date"), tomorrow);

    query.where(criteriaBuilder.and(ownerCondition, dateCondition));


    List<QuestionSet> result = JPA.em().createQuery(query).getResultList();

    return result;

但我得到

play.api.Application$$anon$1: Execution exception[[IllegalArgumentException: Unable to resolve attribute [appointedRepetition.date] against path [null]]]

查看如何将带有子查询的 JPQL 转换为等效的 Criteria API？我在标准 api 部分有几乎相同的代码。

@Entity
@SequenceGenerator(name = "wordlist_seq", sequenceName = "wordlist_seq")
public class QuestionSet {
  @OneToMany(cascade = CascadeType.ALL)
  private List<Question> questions;
      ...
}


@Entity
@SequenceGenerator(name = "question_seq", sequenceName = "question_seq")
@Inheritance(strategy=InheritanceType.TABLE_PER_CLASS)
public abstract class Question{
    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "question_seq")
    private Long id;
        ...
}


    @OneToOne(cascade = CascadeType.ALL)
    private AppointedRepetition appointedRepetition;
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1 回答 1

2

您需要另一个联接，但我不能保证它会起作用，因为实体定义要么丢失要么不完整，并且并非所有关系都按照您的评论中所述进行定义。无论如何，我会试试这个：

Join<Question, AppointedRepetition> repetition = questionJoin.join("appointedRepetition");
Predicate dateCondition = criteriaBuilder.lessThan(repetition.get("date"), tomorrow);

顺便说一句，我看到您正在使用 joda 的 DateTime。我从未将它与 JPA CriteriaBuilder 一起使用，所以我不能保证它有效。

于 2013-06-03T14:21:49.643 回答