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member（这是一个left join）对象上有一个不同的记录，而不是membership。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，是给我成员身份distinct object而不是成员身份distinct object

query.distinct(true);

所以在这里我如何得到成员distinct对象，它又是一个左连接。实体结构如下：

@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。提前谢谢！！！