在這里，我試圖找到所有在我提供的范圍內(nèi)的實(shí)體。我的意思是，如果我給出一個(gè)特定半徑的圓，它必須顯示所有實(shí)體的位置坐標(biāo)位于給定的圓內(nèi)。我正在使用休眠空間來(lái)實(shí)現(xiàn)這一點(diǎn)。但是在JPA存儲(chǔ)庫(kù)接口中獲得提到的錯(cuò)誤。這是 ，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@NoArgsConstructorpublic class Resource extends UUIDEntity2 implements IsResource {    @Type(type = "org.hibernate.spatial.GeometryType")    @OneToOne    private Address address;    /*getters setters*/}Address.java,@Entitypublic 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;}測(cè)試    @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();    }