Spring Data JPA:如何在JPA规范中的左连接字段上获得不同的记录

Question

如何用JPA规范替换下面的Hibernate查询

String hql = "select distinct m " +
                "from Membership ms " +
                "left join ms.member m " +
                "where lower(m.lastName) = :lastName " +
                "and lower(m.firstName) = :firstName " +
                "and m.gender = :gender ";

从这里的主要问题是我想要一个独特的记录成员（这是一个左加入）对象，而不是成员资格。下面是我尝试的：

    @Override
    public Specification<Member> buildSpecification(Membership m) {
        Member member = m.getMember();
        return (root, query, criteriaBuilder) -> {
                PredicateBuilder predicateBuilder = new PredicateBuilder(criteriaBuilder);
                Join<?, ?> memberJoin = root.join("member", JoinType.LEFT);
                List<Predicate> predicates = predicateBuilder
                        .addEqualLowerPredicate(member.getLastName().toLowerCase(), memberJoin.get("lastName"))
                        .addEqualLowerPredicate(member.getFirstName().toLowerCase(), memberJoin.get("firstName"))
                        .addEqualPredicate(member.getGender(), memberJoin.get("gender"))
                        .getPredicates();
                query.distinct(true);
                return criteriaBuilder.and(predicates.toArray(new Predicate[0]));
        };
}

在这里，应用 distinct as true 后，给我成员资格不同对象不是成员不同对象

 query.distinct(true);

所以在这里我如何才能获得成员不同的对象，这又是一个左连接。下面是实体结构：

@Entity
public class Membership {

    private String memberId;
    private Date startDate;
    private Date endDate;
    @ManyToOne
    private Member member;
    .......
    ......
}

@Entity
public class Member {

    private String firstName;
    private String lastName;
    private MemberGender gender;
    .......
    ......
}

我对JPA规范很陌生，任何帮助都非常感谢。提前致谢!!!

Answer 1

AFAIU不可能用Spring Data JPA存储库方法来选择子属性，所以如果您需要查询/存储库根是< code>Membership，那么无论如何都需要一个查询。如果您可以在< code>MemberRepository上定义此方法，这将会起作用。