2017-02-15 64 views
2

我當時在使用AWS託管服務時首先爲我的應用程序使用了Parse,但後來當Parse關閉其服務時,我將其移至back4app。我的登錄和註冊過去與以前的Parse項目一起工作良好,但現在無論何時我使用我的back4app項目登錄或註冊,我都會收到錯誤com.parse.ParseRequest $ ParseRequestException:未授權。我嘗試過使用客戶端密鑰和主密鑰,但他們似乎根本沒有工作。

這裏是我的註冊代碼:

public void clickSignup(View view) { 


    ParseUser.logOut(); 

    loader = (ProgressBar) findViewById(R.id.loadingBar); 

    loader.setVisibility(View.VISIBLE); 

    ParseQuery checkAvailability = ParseUser.getQuery(); 

    textUser = (EditText) findViewById(R.id.textUser); 
    textPass = (EditText) findViewById(R.id.textPass); 

    if (textUser.getText().toString().isEmpty() || textPass.getText().toString().isEmpty()) { 

     loader.setVisibility(View.INVISIBLE); 


     Toaster("Enter a valid username/password"); 



    } else if (textUser.getText().length() < 4){ 

     Toaster("Username must have four or more characters"); 

     loader.setVisibility(View.INVISIBLE); 



    } else if (textPass.getText().length() < 6) { 

     Toaster("Password mush have atleast six or more characters"); 

     loader.setVisibility(View.INVISIBLE); 



    }else{ 

     String user = textUser.getText().toString(); 

     Log.i("user", user); 

     checkAvailability.whereEqualTo("username" , user); 

     checkAvailability.findInBackground(new FindCallback<ParseUser>() { 
      @Override 
      public void done(List<ParseUser> objects, ParseException e) { 

       if (e == null && objects.size() == 1){ 

        loader.setVisibility(View.INVISIBLE); 

        Toaster("This username already exists"); 

       } else if (e == null && objects.size() != 1){ 

        loader.setVisibility(View.INVISIBLE); 

        ParseUser newUser = new ParseUser(); 

        newUser.setUsername(textUser.getText().toString()); 
        newUser.setPassword(textPass.getText().toString()); 

        newUser.signUpInBackground(new SignUpCallback() { 
         @Override 
         public void done(ParseException e) { 

          Log.i("S", "Seucbbkw"); 
          Log.i("e", e.toString()); 

          if (e == null){ 

           Intent intent = new Intent(getApplicationContext(), LoggedIn.class); 
           intent.putExtra("Username",textUser.getText().toString()); 
           startActivity(intent); 

           Toaster("Signup Successful"); 

          } 


         } 
        }); 


       } else { 

        Log.i("Objects", e.toString()); 

       } 



      } 
     }); 


    } 

} 

,這裏是我的登入碼:

public void clickSignin(View view){ 

    ParseUser.logOut(); 

    textUser = (EditText) findViewById(R.id.textUser); 
    textPass = (EditText) findViewById(R.id.textPass); 
    loader.setVisibility(View.VISIBLE); 


    if (textUser.getText().toString().isEmpty() || textPass.getText().toString().isEmpty()) { 

     loader.setVisibility(View.INVISIBLE); 


     Toaster("Enter a valid username/password"); 

    } else { 

     final ParseQuery logInner = ParseUser.getQuery(); 

     logInner.whereEqualTo("username", textUser.getText().toString()); 

     ParseUser.logInInBackground(textUser.getText().toString(), textPass.getText().toString(), new LogInCallback() { 
      @Override 
      public void done(ParseUser user, ParseException e) { 
       if (e == null) { 

        Intent intent = new Intent(getApplicationContext(), LoggedIn.class); 
        intent.putExtra("Username",textUser.getText().toString()); 

        startActivity(intent); 



        Toaster("Login Successful"); 

        loader.setVisibility(View.INVISIBLE); 

       } else { 

        Toaster("Invalid username or password"); 

        Log.i("Error", e.toString()); 

        loader.setVisibility(View.INVISIBLE); 

       } 
      } 
     }); 
    } 
} 

回答

0

(只是爲了測試),我建議你使用的appid太在Back4App連接!例如代碼如下:

<resources> 

<string name="back4app_server_url">https://parseapi.back4app.com/</string> 

<!-- Change the following strings as required --> 
<string name="back4app_app_id">PASTE_YOUR_APPLICATION_ID_HERE</string> 
<string name="back4app_client_key">PASTE_YOUR_CLIENT_KEY_HERE</string> 
<string name="back4app_master_key">PASTE_YOUR_MASTER_KEY_HERE</string> 
<string name="app_name">QuickstartExampleApp</string> 

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