jpa - 如何在内部查询中引用外部查询的成员？

Question

我想在 JPA 2.1 Criteria API 中编写此 SQL 查询:

select * from t_question q
where
(select count(*) from t_question_tag tag 
   where 
       q.question_id  = tag.question_id
       AND tag.tag_id in (18, 1)
) = 2;

我不知道如何在内部查询中引用外部问题成员。

我目前在这一点上:

      CriteriaQuery<Question> cq = criteriaBuilder.createQuery(Question.class);
Root<Question> questions = cq.from(Question.class);
cq.distinct(true);

  Subquery<Long> selectTags = cq.subquery(Long.class);
  Root<QuestionTag> qt = selectTags.from(QuestionTag.class);
  Join<QuestionTag, Question> qtJoin = qt.join("question");
  selectTags
    .select(criteriaBuilder.count(qtJoin))
    .where(
        qt.get("tag").in(filter.getTags())
        );
  cq.where(criteriaBuilder.and(insArray),
      criteriaBuilder.equal(criteriaBuilder.literal(filter.getTags().size()), selectTags));

但是它创建了第二个 Join。 SQL 结果是:

SELECT DISTINCT ...
FROM T_QUESTION question0_
WHERE 1                    =
  (SELECT COUNT(question3_.question_id)
  FROM T_QUESTION_TAG questionta2_
  INNER JOIN T_QUESTION question3_
  ON questionta2_.question_id=question3_.question_id
  WHERE questionta2_.tag_id IN (18));

Answer 1

我希望子查询更像这样

Subquery<Long> selectTags = cq.subquery(Long.class);
Root<QuestionTag> qt = selectTags.from(QuestionTag.class);
selectTags.select(criteriaBuilder.count(qt));
selectTags.where(
        criteriaBuilder.equal(questions.get("id"), qt.get("id")),
        qt.get("tag").in(filter.getTags())
        );

使用来自外部查询的候选项(“问题”)来引用外部查询，并且也不明白您之前为什么要进行连接。我假设“问题”和“标签”中的字段都称为“id”。