Question

我想将{"userid": "userid","pass":"1222"}的POST RAW数据作为用户名和密码发送。我有一个由用户名和密码组成的布局，将作为用户ID和密码获取。我需要帮助才能尝试改装

   // 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
            }
        }

请任何帮助，因为我经常搜索并且没有达成任何帮助

Answer 1

首先：你应该创建你的api界面

      public interface Api
     {
     @Headers({"Accept: application/json"})
     @FormUrlEncoded
     @POST("authentication.php")
     Call<Void> Login(@Field("[email]") String email,
                 @Field("[password]") String password);
     }

第二：在您的活动中，您应该调用您的函数

    void Login(String email, final String password)
    {
    Retrofit retrofit = new Retrofit.Builder()
            .baseUrl("http://mark.journeytech.com.ph/mobile_api/")
            .addConverterFactory(GsonConverterFactory.create())
            .build();
    Apiservice = retrofit.create(Api.class);
    Call<Void> call = service.Login(email, password);
    call.enqueue(new Callback<Void>()
    {
        @Override
        public void onResponse(Call<Void> call, Response<Void> response)
        {
                if (response.isSuccess())
               {

               }
                else
                {
                }
            }
        @Override
        public void onFailure(Call<Void> call, Throwable t)
        {
        }
    });
}

Answer 2

您需要按照以下步骤操作：

网络API接口

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

登录请求模型类

   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
        }
    });
}

在您的活动中改装电话

String baseUrl =＆＃34; http：//mark.journeytech.com.ph/mobile_api/" ;;

NetworkAPI networkAPI;

{{1}}

Android系统。如何使用改造或帮助使用asynctask来发布原始数据

2 个答案: