java - 安卓。如何使用改造发布原始数据或帮助使用 asynctask

Question

我想发送 {"userid": "userid","pass":"1222"} 的 POST RAW 数据作为用户名和密码。我有一种由用户名和密码组成的布局，它将作为用户名和密码获取。我需要帮助来尝试改造

   // Triggers when LOGIN Button clicked
    public void checkLogin(View arg0) {

            // Initialize  AsyncLogin() class with userid and password
            new REST().execute();

    }



    public class REST extends AsyncTask<Void, Void, Void> {
        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            // The Username & Password
            final EditText usr =  (EditText) findViewById(R.id.username);
            userid = (String) usr.getText().toString();
            final EditText pw =  (EditText) findViewById(R.id.password);
            password = (String) pw.getText().toString();
        }

        @Override
        protected Void doInBackground(Void... params) {
            HttpURLConnection urlConnection=null;
            String json = null;

            // -----------------------

            try {
                HttpResponse response;
                JSONObject jsonObject = new JSONObject();
                jsonObject.accumulate("username", usr);
                jsonObject.accumulate("password", password);
                json = jsonObject.toString();
                HttpClient httpClient = new DefaultHttpClient();
                HttpPost httpPost = new HttpPost("http://mark.journeytech.com.ph/mobile_api/authentication.php");
                httpPost.setEntity(new StringEntity(json, "UTF-8"));
                httpPost.setHeader("Content-Type", "application/json");
                httpPost.setHeader("Accept-Encoding", "application/json");
                httpPost.setHeader("Accept-Language", "en-US");
                response = httpClient.execute(httpPost);
                String sresponse = response.getEntity().toString();
                Log.w("QueingSystem", sresponse);
                Log.w("QueingSystem", EntityUtils.toString(response.getEntity()));
            }
            catch (Exception e) {
                Log.d("InputStream", e.getLocalizedMessage());

            } finally {
// nothing to do here

            }
            return null;
        }

        @Override
        protected void onPostExecute(Void result) {
            Toast.makeText(getApplicationContext(), email + " "+ password, Toast.LENGTH_SHORT).show();
            if (result != null) {
                // do something
            } else {
                // error occured
            }
        }

请帮忙，因为我搜索了很多，但没有找到任何东西

Answer 1

您需要执行以下步骤:

1. 网络API接口(interface)

 public interface NetworkAPI {
    @GET("authentication.php")
    @Headers({"Content-Type:application/json; charset=UTF-8"})
    Call<JsonElement> loginRequest(@Body LoginRequest body);
}

登录请求模型类

public class LoginRequest {
    String userid;
    String password;
    public LoginRequest(String userid, String password) {
        this. userid = userid;
        this. pass = pass;
    }
}

在您的 Activity 中进行改造调用

String baseUrl ="http://mark.journeytech.com.ph/mobile_api/";

NetworkAPI networkAPI;

   public void loginRequest(){

    Retrofit retrofit = new Retrofit.Builder()
            .baseUrl(baseUrl)
            .addConverterFactory(GsonConverterFactory.create())
            .build();

    networkAPI = retrofit.create(NetworkAPI.class);

    LoginRequest loginRequest = new LoginRequest(yourusername,yourpassword);

    Call<JsonElement> call = networkAPI.loginRequest(loginRequest);

    call.enqueue(new Callback<JsonElement>() {
        @Override
        public void onResponse(Call<JsonElement> call, Response<JsonElement> response) {
            // success response

        }

        @Override
        public void onFailure(Call<JsonElement> call, Throwable t) {
           // failure response
        }
    });
}