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select ptm.* from ProofTestMaster ptm LEFT JOIN 
ProofTestApplicationMap ptam on ptm.proofTestID = ptam.proofTestID 
LEFT JOIN ProofTestComapartmentMap ptcm on ptm.proofTestID = ptcm.proofTestID 
where (ptam.applicationID = 3 and ptm.isDeleted = 0) or 
(ptcm.compartmentID = 4 and ptm.isDeleted = 0)

映射表在哪里ProofTestApplicationMap，ProofTestComapartmentMap它们在 java 端没有实体。

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1 回答 1

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为您的查询放置适当的缩进总是更好。优点是：

  • 你会很容易理解查询
  • 其他人也会很快明白的。

所以我为你做了这个。现在查询看起来像：

select 
    ptm.* 
from ProofTestMaster ptm 
    LEFT JOIN ProofTestApplicationMap ptam on ptm.proofTestID = ptam.proofTestID 
    LEFT JOIN ProofTestComapartmentMap ptcm on ptm.proofTestID = ptcm.proofTestID 
where 
    (ptam.applicationID = 3 and ptm.isDeleted = 0) or 
    (ptcm.compartmentID = 4 and ptm.isDeleted = 0);

下面是 CriteriaBuilder 和静态元模型的实现：

CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<Long> criteriaQuery = criteriaBuilder.createQuery(Long.class);

Root<ProofTestMaster> mainRoot = criteriaQuery.from(ProofTestMaster.class);

Join<ProofTestMaster, ProofTestApplicationMap> firstJoin = mainRoot.join(ProofTestMaster_.proofTestID, JoinType.LEFT);
Join<ProofTestMaster, ProofTestComapartmentMap> secondJoin = mainRoot.join(ProofTestMaster_.proofTestID, JoinType.LEFT);

Predicate p1 = criteriaBuilder.equal(firstJoin.get(ProofTestApplicationMap_.applicationID),3);
Predicate p2 = criteriaBuilder.equal(mainRoot.get(ProofTestMaster_.isDeleted),0);
Predicate p3 = criteriaBuilder.equal(secondJoin.get(ProofTestComapartmentMap_.compartmentID), 4);
Predicate p4 = criteriaBuilder.and(p1,p2);
Predicate p5 = criteriaBuilder.and(p3,p2);
Predicate p6 = criteriaBuilder.or(p4,p5);

criteriaQuery.where(p6);

criteriaQuery.select(criteriaBuilder.count(mainRoot));

Long count = entityManager.createQuery(criteriaQuery).getSingleResult();

如果你看到上面的代码，总共有 6 个 Predicates，可以放入 List。但为了您的理解，我保留了它。

让我知道它是否对您有帮助。感谢和快乐的编码。

于 2013-08-19T12:00:29.887 回答