在这里,我试图找到所有在我提供的范围内的实体。我的意思是,如果我给出一个特定半径的圆,它必须显示所有实体的位置坐标位于给定的圆内。
我正在使用休眠空间来实现这一点。但是在JPA存储库接口中获得提到的错误。
这是 ,pom.xml
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-entitymanager</artifactId>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-spatial</artifactId>
<version>5.2.12.Final</version>
</dependency>
<dependency>
<groupId>org.opengeo</groupId>
<artifactId>geodb</artifactId>
<version>${project.version}</version>
</dependency>
<dependency>
<groupId>mysql</groupId>
<artifactId>mysql-connector-java</artifactId>
<version>6.0.6</version>
</dependency>
Jpa Repository,
public interface ResourceRepository extends ExtendedJpaRepository<Resource, String> {
@Query(value = "select resource from Resource resource where within(resource.address.location, :circle) = true")
List<Resource> test(@Param("circle") Geometry circle);
}
Resource.java,
@Entity
@NoArgsConstructor
public class Resource extends UUIDEntity2 implements IsResource {
@Type(type = "org.hibernate.spatial.GeometryType")
@OneToOne
private Address address;
/*getters setters*/
}
Address.java,
@Entity
public class Address extends UUIDEntity2 implements HasEmailAddress, HasLocation {
@Embedded
@Column(columnDefinition = "point")
private Location location;
/*getters setters*/
}
location.java,
@Embeddable
@Value(staticConstructor = "of")
@RequiredArgsConstructor(staticName = "of")
public class Location implements Serializable {
@Column(nullable = true)
private Double lat;
@Column(nullable = true)
private Double lon;
}
测试
@Inject
private ResourceRepository resourceRepository;
public Geometry createCircle(double x, double y, double radius) {
GeometricShapeFactory shapeFactory = new GeometricShapeFactory();
shapeFactory.setNumPoints(32);
shapeFactory.setCentre(new Coordinate(x, y));
shapeFactory.setSize(radius * 2);
return shapeFactory.createCircle();
}
森栏
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