什么是Android及其使用的严格性

Question

I am new to android , i am following this tutorial, i have found the code below , there he is converting json string to StringEntity. correct me if i am wrong StringEntity is used to pass the data,Headers like Accept,Content-type to Server.

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

        // 2. make POST request to the given URL
        HttpPost httpPost = new HttpPost(url);



        String json = "";

        // 3. build jsonObject
        JSONObject jsonObject = new JSONObject();
        jsonObject.accumulate("name", person.getName());
        jsonObject.accumulate("country", person.getCountry());
        jsonObject.accumulate("twitter", person.getTwitter());

        // 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);

        // 5. set json to StringEntity
        StringEntity se = new StringEntity(json);

        // 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();
.
.
.

and how do i get the data in the servlet/jsp ? Should i use getStream() or request.getParameter()

Answer 1

An entity whose content is retrieved from a string.

StringEntity is the raw data that you send in the request.

Server communicate using JSON, JSON string can be sent via StringEntity and server can get it in the request body, parse it and generate appropriate response.

we set all our unicode style,content type in this only

StringEntity se = new StringEntity(str,"UTF-8");
    se.setContentType("application/json");
    httpPost.setEntity(se); 

For more help u can take reference this http://developer.android.com/reference/org/apache/http/entity/StringEntity.html

As per your requirement I edit this for post method

HttpPost httpPost = new HttpPost(url_src);
HttpParams httpParameters = new BasicHttpParams();
httpclient.setParams(httpParameters);
StringEntity se = new StringEntity(str,"UTF-8");
se.setContentType("application/json");
httpPost.setEntity(se); 



try
{
    response = httpclient.execute(httpPost);

    StatusLine statusLine = response.getStatusLine();
    int statusCode = statusLine.getStatusCode();


    if(statusCode==200)
    {
        entity = response.getEntity();
        String responseText = EntityUtils.toString(entity);
        System.out.println("The response is" + responseText);   

    }
    else
    {
        System.out.println("error");;
    }
}
catch(Exception e)
{

    e.printStackTrace();

}

Answer 2

StringEntity is the raw data that you send in the request.

Most of the server communicate using JSON, JSON string can be sent via StringEntity and server can get it in the request body, parse it and generate appropriate response.

Accept,Content-type etc. are sent as the header of the request but StringEntity is content of it.

Header is not passed in StringEntity .

Answer 3

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)Example: 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)Example : 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)