顺便说一句,我在 Netbeans/Glashfish 中与Jackson在3 个步骤中解决了同样的问题。
1)要求:
我使用的一些罐子:
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
如果我错过了上面的任何 jar,你可以从这里的 Maven 下载http://mvnrepository.com/artifact/com.fasterxml.jackson.core
2)您发送帖子的Java类。首先,使用 Jackson 将实体用户转换为 Json,然后将其发送到您的 Rest 类。
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 你想要接收实体 JPA/Hibernate 的地方。在这里,您可以使用 MediaType.APPLICATION_JSON 以这种方式接收实体:
""id": 99 ,"usuarioPadre":null,"昵称":null,"clave":null,"nombre":null,"apellidos":null,"isLoginWeb":null,"isLoginMovil":null," estado":null,"correoElectronic":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
}
..
一些提示:如果您在使用此代码后运行网络有问题可能是因为 @Consumes in XML
...您必须将其设置为@Consumes(MediaType.APPLICATION_JSON)