org.hibernate.engine.jdbc.spi.SqlExceptionHelper - 找不到函数“WITHIN”;SQL语句。

3

我正在尝试找到所有在我提供的范围内的实体。我的意思是,如果我给定一个特定半径的圆,它将必须显示所有位置坐标位于给定圆内的实体。

我使用hibernate-spatial来实现这一点。但在JPA Repository接口中遇到了上述错误。

这是pom.xml文件:

<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仓库,

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();
    }

    @Test
    public void geometry(){
        Geometry m = createCircle(0.0, 0.0, 5);
        List<Resource> resources = resourceRepository.test(m);
    }

application.properties,

hibernate.dialect=org.hibernate.spatial.dialect.mysql.MySQL56SpatialDialect

注意:这里未显示实体的所有属性。 我所遵循的参考资料为:Hibernate-Spatial
1个回答

1
听起来你没有配置MySQL的SpatialDialect。你能确认是否有以下这行代码吗?
hibernate.dialect=org.hibernate.spatial.dialect.mysql.MySQL56SpatialDialect

hibernate.properties文件中。
您还可以检查日志以查看Hibernate实际使用的方言是什么。它应该在名称中具有“Spatial”,以便空间函数对Hibernate可用。

我已经在application.properties文件中添加了它。但是日志显示它正在使用'org.hibernate.dialect.H2Dialect'。那么该如何配置呢? - Chinmoy Acharjee

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