Spring Data存储库和带有Geometry返回类型的本机查询

Question

我正在创建一个使用一些空间查询的项目。我使用带有Spring数据存储库的Spring引导和带有PostGIS扩展的PostgreSQL作为数据库。

我创建了这个存储库：

import com.vividsolutions.jts.geom.Geometry;
import org.springframework.data.jpa.repository.Query;
import org.springframework.data.repository.CrudRepository;
import org.springframework.data.repository.query.Param;
import org.springframework.stereotype.Repository;

@Repository
public interface AreaRepository extends CrudRepository<Area, Long> {

    /*
      extra queries for Area here
    */

    @Query(value="select st_intersection(" +
                ":base_layer ," +
                ":filter_layer" +
                ")", nativeQuery = true)
    Geometry geometryIntersectGeometry(@Param("base_layer") Geometry baseGeometry,@Param("filter_layer") Geometry filterGeometry);

}

它包含对Area实体的一些查询。我还想使用一些PostGIS函数来进行一些计算，所以我创建了geometryIntersectGeometry来从PostGis调用st_intersection函数，这应该返回一个几何。

我在设置中将hibernate方言设置为PostGIS：

spring.datasource.driverClassName=org.postgresql.Driver
spring.datasource.url=jdbc:postgresql://localhost:5432/test_db
spring.datasource.username=postgres
spring.datasource.password=postgres
spring.jpa.properties.hibernate.dialect = org.hibernate.spatial.dialect.postgis.PostgisDialect

我有hibernate空间的依赖项：

        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-data-jpa</artifactId>
        </dependency>
        <dependency>
            <groupId>org.postgresql</groupId>
            <artifactId>postgresql</artifactId>
            <scope>runtime</scope>
        </dependency>
        <dependency>
            <groupId>org.hibernate</groupId>
            <artifactId>hibernate-spatial</artifactId>
            <version>${hibernate.version}</version>
        </dependency>
        ...

调用geometryIntersectGeometry函数会导致错误：

No Dialect mapping for JDBC type: 1111; nested exception is org.hibernate.MappingException: No Dialect mapping for JDBC type: 1111,{}

如何告诉JPA / Spring Data将几何（PostGIS类型）响应映射到Geometry（com.vividsolutions.jts.geom.Geometry）对象？

Answer 1

你添加了Hibernate Spatial吗？

<dependency>
    <groupId>org.hibernate</groupId>
    <artifactId>hibernate-spatial</artifactId>
    <version>${hibernate.version}</version>
</dependency>

这支持GIS数据：http://docs.jboss.org/hibernate/orm/5.2/userguide/html_single/Hibernate_User_Guide.html#spatial

Answer 2

管理通过编写存储库的custom implementation来修复它并注册类型（thx Simon Martinelli）

库：

import org.springframework.data.jpa.repository.Query;
import org.springframework.data.repository.CrudRepository;
import org.springframework.data.repository.query.Param;
import org.springframework.stereotype.Repository;

@Repository
public interface AreaRepository extends CrudRepository<Area, Long>, AreaGisRepository {

    /*
      extra queries for Area here
    */

}

接口：

import com.vividsolutions.jts.geom.Geometry;

public interface AreaGisRepository {

    Geometry geometryIntersectGeometry(Geometry baseGeometry, Geometry filterGeometry);
}

和实施：

import com.vividsolutions.jts.geom.Geometry;
import org.hibernate.spatial.JTSGeometryType;
import org.hibernate.spatial.dialect.postgis.PGGeometryTypeDescriptor;
import org.springframework.beans.factory.annotation.Autowired;

import javax.persistence.EntityManager;

public class AreaGisRepositoryImpl implements AreaGisRepository {

    private EntityManager entityManager;

    @Autowired
    public AreaGisRepositoryImpl(EntityManager entityManager) {
        this.entityManager = entityManager;
    }

    @Override
    public Geometry geometryIntersectGeometry(Geometry baseGeometry, Geometry filterGeometry) {
        return (Geometry) entityManager
                .createNativeQuery(
                        "select st_intersection(:base_layer , :filter_layer) as geom")
                .setParameter("base_layer", baseGeometry)
                .setParameter("filter_layer", filterGeometry)
                .unwrap(org.hibernate.query.NativeQuery.class)
                .addScalar("geom", new JTSGeometryType(PGGeometryTypeDescriptor.INSTANCE))
                .getSingleResult();
    }

}

它工作得很完美，但我现在对Postgis有一个硬编码依赖（不太可能我们会使用其他东西，但......）

Answer 3

我认为有一种更简单的方法可以完成这项工作。 Hibernate Spatial注册了许多空间函数，以便在HQL / JQL中使用。所以以下应该有效

@Query(value="select intersection(" +
            ":base_layer ," +
            ":filter_layer" +
            ")")
Geometry geometryIntersectGeometry(@Param("base_layer") Geometry baseGeometry,@Param("filter_layer") Geometry filterGeometry);

有关Spatial Dialects中可用功能的列表，请参阅documentation。