我的应用程序中有以下实体：
MemberFamilyAdvertismentAddress
在Member实体中：
@OneToOne(cascade=CascadeType.ALL)
private Address address;
...
@OneToMany(fetch = FetchType.LAZY, mappedBy = "member")
private List<Advertisement> advertisements;
在Advertisement实体中：
@NotNull
@ManyToOne(fetch = FetchType.LAZY)
private Member member;
完整Address实体：
@Entity
public class Address {
private String formattedAddress;
private double latitude;
private double longitude;
}
我正在尝试查找其成员的地址在所需地址 20KM 范围内的所有 FamilyAdvertisement 实例。
这是我想出的：
QFamilyAdvertisement qFamilyAdvertisement = QFamilyAdvertisement.familyAdvertisement;
NumberPath<Double> lat = qFamilyAdvertisement.member.address.latitude;//NPE
NumberPath<Double> lng = qFamilyAdvertisement.member.address.longitude;
NumberPath<Double> distance = null;
NumberExpression<Double> formula =
(acos(cos(radians(Expressions.constant(requiredAddress.getLatitude())))
.multiply(cos(radians(lat))
.multiply(cos(radians(lng).subtract(radians(Expressions.constant(requiredAddress.getLongitude())))
.add(sin(radians(Expressions.constant(requiredAddress.getLatitude())))
.multiply(sin(radians(lat))))))))
.multiply(Expressions.constant(6371)));
List<FamilyAdvertisement> foundFamilyAdvertisements = from(qFamilyAdvertisement.member.address).where(formula.as(distance).lt(20)).list(qFamilyAdvertisement);
但是，当我不断获得 NPE 时，似乎我错误地使用了 NumberPath 类。任何人都可以帮我正确查询吗？
编辑：我已将我的 FamilyAdvertisement 实体更改如下：
@NotNull
@ManyToOne(fetch = FetchType.LAZY)
@QueryInit("address")
private Member member;
我现在得到以下异常：
java.lang.IllegalArgumentException: Only root paths are allowed for joins : familyAdvertisement.member.address
com.mysema.query.DefaultQueryMetadata.ensureRoot(DefaultQueryMetadata.java:208)
com.mysema.query.DefaultQueryMetadata.validateJoin(DefaultQueryMetadata.java:132)
com.mysema.query.DefaultQueryMetadata.addJoin(DefaultQueryMetadata.java:118)
com.mysema.query.DefaultQueryMetadata.addJoin(DefaultQueryMetadata.java:110)
com.mysema.query.support.QueryMixin.from(QueryMixin.java:161)
com.mysema.query.jpa.JPQLQueryBase.from(JPQLQueryBase.java:96)
com.mysema.query.jpa.impl.JPAQuery.from(JPAQuery.java:30)
org.springframework.data.jpa.repository.support.Querydsl.createQuery(Querydsl.java:88)
org.springframework.data.jpa.repository.support.QueryDslRepositorySupport.from(QueryDslRepositorySupport.java:94)
com.bignibou.repository.FamilyAdvertisementRepositoryImpl.performFamilyAdvertisementSearch(FamilyAdvertisementRepositoryImpl.java:64)
第 64 行是这一行：
List<FamilyAdvertisement> foundFamilyAdvertisements = from(qFamilyAdvertisement.member.address).where(formula.as(distance).lt(20)).list(qFamilyAdvertisement);
任何线索现在出了什么问题？
edit2：我忘了提到FamilyAdvertisement扩展Advertisement和member变量在Advertisement.
edit3：这是我试图用 QueryDSL 重现的 SQL：
select * from family_advertisement a inner join member m
on a.member = m.id
where m.address
in (
SELECT id
FROM address where
6371 *
acos( cos( radians(48.8558966) )
* cos( radians( latitude ) )
* cos( radians( longitude ) - radians(2.3622728) )
+ sin( radians(48.8558966) )
* sin( radians( latitude ) )
) < 20);
我尝试过这样的事情：
List<FamilyAdvertisement> foundFamilyAdvertisements = from(qFamilyAdvertisement).where(qFamilyAdvertisement.member.address.in(
new JPASubQuery().from(QAddress.address).where(formula.lt(20)))
).list(qFamilyAdvertisement);
上面给出了公式，但我不确定如何在 QueryDSL 中表达不相关的子查询，尤其是上面的 in 运算符似乎有问题......
编辑4：
以下子查询现在可以工作：
List<FamilyAdvertisement> foundFamilyAdvertisements =
from(qFamilyAdvertisement).where(qFamilyAdvertisement.member.address.in(new JPASubQuery().from(QAddress.address).where(formula.lt(20)).list(QAddress.address))).list(qFamilyAdvertisement);