How to get geoJSON from spring REST Controller?

问题:

I am developing a GIS application with java(Spring-4.1.5 + Hibernate-4.3.8) and OpenLayers. For this project I use GeoTools-13RC, HibernateSptial-4.3, jts-1.13 and jackson-2.5. In this project, I have a layer in client side and in server, I save the features of this layer in a class. I defined the class below:

@Entity
@Table(name="MyPoint")
public class MyPoint
{
    @id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long Id;

    @Column
    private String name;

    @Type(type = "org.hibernate.spatial.GeometryType")
    @Column(name = "the_geom")
    private Point geometry;

    /*
    *
    Getter and Setter
    *
    */
}

In start up of application, I need to init the layer in client side. for this, I need return from server side a json string to client for this layer. I don't want to use ST_AsGeoJson or other matches. I use Spring REST controller for returning my Entity.

What do I do?

回答1:

You can use jackson-datatype-jts from this link.
Its integration with spring is as below:

dispather-servlet.xml:

<mvc:annotation-driven >
    <mvc:message-converters>
        <bean class="org.springframework.http.converter.json.MappingJackson2HttpMessageConverter">
          <property name="objectMapper" ref="customObjectMapper"/>
      </bean>
    </mvc:message-converters>
</mvc:annotation-driven>
<bean id="customObjectMapper" class="my.server.util.CustomObjectMapper"/><!--custom jackson objectMapper -->

my.server.util.CustomObjectMapper:

package my.server.util;

import com.bedatadriven.jackson.datatype.jts.JtsModule;
import com.fasterxml.jackson.core.JsonFactory;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.deser.DefaultDeserializationContext;
import com.fasterxml.jackson.databind.ser.DefaultSerializerProvider;
import org.springframework.beans.factory.InitializingBean;

/**
 *
 * @author dariush
 */
public class CustomObjectMapper extends ObjectMapper implements InitializingBean{

   public CustomObjectMapper() {
   }

   public CustomObjectMapper(JsonFactory jf) {
       super(jf);
   }

   public CustomObjectMapper(ObjectMapper src) {
       super(src);
   }

   public CustomObjectMapper(JsonFactory jf, DefaultSerializerProvider sp, DefaultDeserializationContext dc) {
       super(jf, sp, dc);
   }

   @Override
   public void afterPropertiesSet() throws Exception {
       this.registerModule(new JtsModule());
   }

}

Then in the rest controller if you return a Geometry, you can get its geojson.

回答2:

Returning a Response to the Client

You will need to create a resource to expose to your client. There is some good Spring documentation on this topic, and a couple of different ways to do it, but essentially something like this:

@Component
@Path("/my_points")
public class MyPoints {

    private PointService pointService;

    @GET
    @Path("{pointId}")
    public Response getPoint(@PathParam("pointId") String pointId) {
        return Response.ok(pointService.getById(pointId)).build();
    }
}

Building JSON

You should use Jackson to build your JSON. If you build a Spring resource then Jackson will likely be used by default when constructing a response. To give you an idea of how to translate an object to JSON manually:

@Test
public void serializingAnObjectToJson() throws Exception {
    // Create a mapper that translates objects to json/xml/etc using Jackson.
    ObjectMapper om = new ObjectMapper();

    MyPoint point = new MyPoint(223l, "Superman");

    // Creates a JSON representation of the object
    String json = om.writeValueAsString(point);

    // Create a JAX-RS response with the JSON as the body
    Response response = Response.ok(json).build();
}