java Spring数据JPA：如何在JPA规范的左连接字段中获得不同的记录

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 ";

这里的主要问题是，我希望在成员（这是一个左连接）对象上有一个清晰的记录，而不是成员。以下是我尝试过的：

    @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，在应用distinct之后，true为我提供了成员身份distinct对象而不是成员身份distinct对象

 query.distinct(true);

所以在这里，我怎样才能得到成员distinct对象，它又是一个左连接。以下是实体结构：

@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规范非常陌生，非常感谢您的帮助。提前感谢