Question

我的申请中有以下实体：

Member
FamilyAdvertisment
Address

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在所需地址的20KM范围内具有地址的所有FamilyAdvertisement实例。

以下是我的想法：

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，我似乎错误地使用了NumberPath类，因为我一直在使用NPE。任何人都可以帮我解决问题吗？

编辑：我已更改了FamilyAdvertisement实体，如下所示：

@NotNull
@ManyToOne(fetch = FetchType.LAZY)
@QueryInit("address")
private Member member;

我现在得到以下例外：

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行就是这个：

List<FamilyAdvertisement> foundFamilyAdvertisements = from(qFamilyAdvertisement.member.address).where(formula.as(distance).lt(20)).list(qFamilyAdvertisement);

现在有什么问题吗？

edit2 ：我忘了提及FamilyAdvertisement延伸Advertisement并且member变量位于Advertisement。

edit3 ：这是我尝试使用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中表达一个不相关的子查询，特别是上面的in运算符似乎有问题......

edit4 ：

现在可以使用以下子查询：

List<FamilyAdvertisement> foundFamilyAdvertisements = 
        from(qFamilyAdvertisement).where(qFamilyAdvertisement.member.address.in(new JPASubQuery().from(QAddress.address).where(formula.lt(20)).list(QAddress.address))).list(qFamilyAdvertisement);

Answer 1

此路径太长，无法进行急切初始化

qFamilyAdvertisement.member.address.latitude;

请在此处阅读有关Querydsl http://www.querydsl.com/static/querydsl/3.1.0/reference/html/ch03s04.html#d0e1699

中路径初始化的更多信息

使用QueryDSL查询数据库中的近距离点

1 个答案: