4

我被困在一个查询中,我正在使用 Spring Data JPA、JpaRepository 和 JpaSpecificationExecutor。所以我必须使用Predicates from CriteriaBuilder

目前我只是这样做:

Specifications<MyEntity> spec = Specifications.where(MySpec.isTrue());
List<MyEntity> myEntities = myRepository.findAll(spec);

其中 MySpec.isTrue() 是:

public static Specification<MyEntity> isTrue() {
    return new Specification<MyEntity>() {
        @Override
        public Predicate toPredicate(Root<MyEntity> root, CriteriaQuery<?> query, CriteriaBuilder cb) {
            SetJoin<MyEntity, JoinEntity> j = root.join(MyEntity.joinEntities, JoinType.LEFT);  
            return cb.isTrue(j.get(JoinEntity_.attr));
        }
    };
}

所以这当然会导致这个 SQL:

SELECT e.* FROM MyEntity e
  LEFT OUTER JOIN JoinEntity j ON j.myEntityId = e.id
  WHERE j.attr = true

但我只对独特的MyEntitys 集感兴趣。所以在 JPQL 我会写:

SELECT DISTINCT(e) FROM MyEntity e
  LEFT JOIN e.joinEntities j
  WHERE j.attr = true

现在我的解决方案是:

List<MyEntity> myEntities = myRepository.findAll(spec);
Set<MyEntity> entitiesSet = new HashSet<MyEntity>(myEntities);

一定有更好的方法;)

它如何与CriteriaBuilder(和JpaSpecificationExecutor)一起使用?


解决方案:

第一个想法很简单:

public static Specification<MyEntity> isTrue() {
    return new Specification<MyEntity>() {
        @Override
        public Predicate toPredicate(Root<MyEntity> root, CriteriaQuery<?> query, CriteriaBuilder cb) {
            SetJoin<MyEntity, JoinEntity> j = root.join(MyEntity.joinEntities, JoinType.LEFT);
            query.distinct(true); // <<-- HERE  
            return cb.isTrue(j.get(JoinEntity_.attr));
        }
    };
}

这行得通,但有点破坏了那些小规格部件的感觉。所以我想出了一个使用子查询的解决方案。这可能需要一些额外的时间,但对我来说这并不重要:

public static Specification<MyEntity> isTrue() {
    return new Specification<MyEntity>() {
        @Override
        public Predicate toPredicate(Root<MyEntity> root, CriteriaQuery<?> query, CriteriaBuilder cb) {
            Subquery<JoinEntity> subquery = query.subquery(JoinEntity.class);
            Root<JoinEntity> subRoot = subquery.from(JoinEntity.class);
            subquery.select(subRoot);
            subquery.where(cb.isTrue(subRoot.get(JoinEntity_.attr)));
            subquery.groupBy(subRoot.get(JoinEntity_.myEntity));

            return cb.exists(subquery);
        }
    };
}
4

1 回答 1

4

您正在寻找CriteriaQuery.distinct(true)

于 2013-07-18T17:15:56.103 回答