Passing an object to a REST Web Service using Jersey

Question

I have a simple WS that is a @PUT and takes in an object

@Path(\"test\")
public class Test {

    @PUT
    @Path(\"{nid}\"}
    @Consumes(\"appl         


        

Answer 1

I had the same problem I solved in 3 Steps with Jackson in Netbeans/Glashfish btw.

1)Requirements :

some of the Jars I used :

 commons-codec-1.10.jar
 commons-logging-1.2.jar
 log4j-1.2.17.jar
 httpcore-4.4.4.jar
 jackson-jaxrs-json-provider-2.6.4.jar
 avalon-logkit-2.2.1.jar
 javax.servlet-api-4.0.0-b01.jar
 httpclient-4.5.1.jar
 jackson-jaxrs-json-provider-2.6.4.jar
 jackson-databind-2.7.0-rc1.jar
 jackson-annotations-2.7.0-rc1.jar
 jackson-core-2.7.0-rc1.jar

If I missed any of the jar above , you can download from Maven here http://mvnrepository.com/artifact/com.fasterxml.jackson.core

2)Java Class where you send your Post. First ,Convert with Jackson the Entity User to Json and then send it to your Rest Class.

import com.fasterxml.jackson.databind.ObjectMapper;
import ht.gouv.mtptc.siiv.model.seguridad.Usuario;
import java.io.IOException;
import java.io.UnsupportedEncodingException;
import org.apache.http.HttpResponse;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.simple.JSONObject;


public class PostRest {


    public static void main(String args[]) throws UnsupportedEncodingException, IOException {

         // 1. create HttpClient
        DefaultHttpClient httpclient = new DefaultHttpClient();

        // 2. make POST request to the given URL
        HttpPost httpPost 
        = new HttpPost("http://localhost:8083/i360/rest/seguridad/obtenerEntidad");


        String json = "";
        Usuario u = new Usuario();
        u.setId(99L);

        // 3. build jsonObject
        JSONObject jsonObject = new JSONObject();
        jsonObject.put("id", u.getId());

        // 4. convert JSONObject to JSON to String
        //json = jsonObject.toString();

        // ** Alternative way to convert Person object to JSON string usin Jackson Lib
         //ObjectMapper mapper = new ObjectMapper();
         //json = mapper.writeValueAsString(person);
        ObjectMapper mapper = new ObjectMapper();
       json = mapper.writeValueAsString(u);

        // 5. set json to StringEntity
        StringEntity se = new StringEntity(json,"UTF-8");


        // 6. set httpPost Entity
        httpPost.setEntity(se);

        // 7. Set some headers to inform server about the type of the content   
        httpPost.setHeader("Accept", "application/json");
        httpPost.setHeader("Content-type", "application/json");

        // 8. Execute POST request to the given URL
        HttpResponse httpResponse = httpclient.execute(httpPost);

        // 9. receive response as inputStream
        //inputStream = httpResponse.getEntity().getContent();

}


}

3)Java Class Rest where you want to receive the Entity JPA/Hibernate . Here with your MediaType.APPLICATION_JSON you recieve the Entity in this way :

""id":99,"usuarioPadre":null,"nickname":null,"clave":null,"nombre":null,"apellidos":null,"isLoginWeb":null,"isLoginMovil":null,"estado":null,"correoElectronico":null,"imagePerfil":null,"perfil":null,"urlCambioClave":null,"telefono":null,"celular":null,"isFree":null,"proyectoUsuarioList":null,"cuentaActiva":null,"keyUser":null,"isCambiaPassword":null,"videoList":null,"idSocial":null,"tipoSocial":null,"idPlanActivo":null,"cantidadMbContratado":null,"cantidadMbConsumido":null,"cuotaMb":null,"fechaInicio":null,"fechaFin":null}"

    import javax.ws.rs.Consumes;
    import javax.ws.rs.GET;
    import javax.ws.rs.POST;

    import javax.ws.rs.Path;
    import javax.ws.rs.PathParam;
    import javax.ws.rs.Produces;
    import javax.ws.rs.core.MediaType;
    import org.json.simple.JSONArray;

    import org.json.simple.JSONObject;
    import org.apache.log4j.Logger;

    @Path("/seguridad")
    public class SeguridadRest implements Serializable {



       @POST
        @Path("obtenerEntidad")
        @Consumes(MediaType.APPLICATION_JSON)
        public JSONArray obtenerEntidad(Usuario u) {
            JSONArray array = new JSONArray();
            LOG.fatal(">>>Finally this is my entity(JPA/Hibernate) which 
will print the ID 99 as showed above :" + u.toString());
            return array;//this is empty

        }
       ..

Some tips : If you have problem with running the web after using this code may be because of the @Consumes in XML ... you must set it as @Consumes(MediaType.APPLICATION_JSON)