android應用程序登錄數據沒有從我的應用程序傳遞到我的php文件

Question

class Login extends Activity implements OnClickListener{ 
    private EditText user, pass; 

    private ProgressDialog pDialog; 

    JSONParser jsonParser = new JSONParser(); 
    private static final String LOGIN_URL = "http://ridgepvt.com/android/loginapp.php"; 
    private static final String TAG_SUCCESS = "success"; private static final String TAG_MESSAGE = "message"; 
    @Override 
    protected void onCreate(Bundle savedInstanceState) { 
     Button bLogin; 
     super.onCreate(savedInstanceState); 
     setContentView(R.layout.login); 
     user = (EditText)findViewById(R.id.username); 
     pass = (EditText)findViewById(R.id.password); 
     bLogin = (Button)findViewById(R.id.login); 
     bLogin.setOnClickListener(this); 
    } 

    @Override 
    public void onClick(View v) { 
     switch (v.getId()) { 
      case R.id.login: 
       String username1 = user.getText().toString(); 
       String password1 = pass.getText().toString(); 
       new AttemptLogin(username1,password1).execute(); 

      default: 
       break; 
     } 
    } 

    class AttemptLogin extends AsyncTask<String, String, String> { 
     boolean failure = false; 
     private String username; 
     private String password; 
     public AttemptLogin(String username1, String password1) { 
      super(); 
      username = username; 
      password = password; 
     } 

     @Override 

     protected void onPreExecute() { 
      super.onPreExecute(); 
      pDialog = new ProgressDialog(Login.this); 
      pDialog.setMessage("Attempting for login..."); 
      pDialog.setIndeterminate(false); 
      pDialog.setCancelable(true); 
      pDialog.show(); 

     } 

     @Override 
     protected String doInBackground(String... args) { 
      int success; 


      try { 
       List<NameValuePair> params = new ArrayList<NameValuePair>(); 
       params.add(new BasicNameValuePair("username", username)); 
       params.add(new BasicNameValuePair("password", password)); 

       Log.d("request!", "starting"); 
       JSONObject json = jsonParser.makeHttpRequest(
         LOGIN_URL, "POST", params); 

       Log.d("Login attempt", json.toString()); 
       success = json.getInt(TAG_SUCCESS); 
       if (success == 1) 
       { Log.d("Successfully Login!", json.toString()); 
        Intent ii = new Intent(Login.this,OtherActivity.class); 

        startActivity(ii); 
        finish(); 
        return json.getString(TAG_MESSAGE); 
       }else{ 
        return json.getString(TAG_MESSAGE); 
       } 
      } 
      catch (JSONException e) 
      { e.printStackTrace(); } 

      return null; 
     } 
     protected void onPostExecute(String message) 
     { pDialog.dismiss(); 
      if (message != null) 
      { 
       Toast.makeText(Login.this, message, Toast.LENGTH_LONG).show(); 
      } 
     } 
    } 
} 

錯誤是邏輯它不會將我的字符串數據傳遞到PHP文件。錯誤是COMING當我在後臺使用GET.TEXT功能，但我想通過重寫構造函數和傳遞數據作爲參數在onclick功能

Answer 1

在構造函數中使用的是

public AttemptLogin(String username1, String password1) { 
     super(); 
     username = username; 
     password = password; 
    } 

解決呢

在doInBackground方法，而不是使用

public AttemptLogin(String username1, String password1) { 
     super(); 
     username = username1; 
     password = password1; 
    }