[英]Spring Data JPA: How to get distinct record on LEFT JOIN field in JPA Specification
如何替换下面的 Hibernate 查询 JPA 规范
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想要在Member (这是一个 LEFT Join) object 而不是Membership上有一个不同的记录。 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]));
};
}
在这里应用distinct as true后给我 Membership distinct object not Member distinct object
query.distinct(true);
所以在这里我怎样才能得到成员不同的 object ,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 规范非常陌生,非常感谢任何帮助。 提前致谢!!!
AFAIU 不可能将 select 子属性与 Spring 数据 JPA 存储库方法联系起来,因此无论如何您都需要一个查询Membership
才能查询/repository。 如果您可以在MemberRepository
上定义此方法,这将起作用。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.