android登錄初學者類

Question

我正在學習android ..以及我在這裏添加了代碼catchmayuri.blogspot.in/2010/12/sample-android-login-application.html我一直在執行該代碼的應用程序沒有運行。 ..

<ImageButton android:id="@+id/buttonLogin" 
    android:layout_width="140dp" 
    android:layout_height="50dp" 
    android:background="@drawable/login5" 
    /> 

這是XML 的代碼如下因素是mainAcivity.java

protected void onCreate(Bundle savedInstanceState) { 

    super.onCreate(savedInstanceState); 
    setContentView(R.layout.activity_main); 

    unameEdit = (EditText) findViewById(R.id.txtUsername); 
    passEdit = (EditText) findViewById(R.id.txtPassword); 
    loginButton = (Button) findViewById(R.id.buttonLogin); 

    System.out.println("button login obtaied"); 
    loginButton.setOnClickListener(new OnClickListener() { 

     @Override 
     public void onClick(View v) { 
      user = unameEdit.getText().toString().trim(); 
      pass = passEdit.getText().toString().trim(); 
      if (user.equalsIgnoreCase(pass)) { 
       Toast.makeText(MainActivity.this, "Login Successfull", 
         Toast.LENGTH_LONG).show(); 
       // setContentView(R.layout.secondpage); Call Intent 

       Intent i = new Intent(MainActivity.this, SecondClass.class); 
       startActivity(i); 
      } else { 
       Toast.makeText(MainActivity.this, 
         "Check the Username/password", Toast.LENGTH_LONG) 
         .show(); 
      } 

     } 
    }); 
} 
} 

apk文件dosent運行的代碼......我到底做錯了什麼??? ？

Answer 1

您得到ClassCastExeption在loginButton = (Button) findViewById(R.id.buttonLogin);。

，如果你正在學習如何實現登錄，想知道有關登錄

的過程中嘗試這種

if(getuname.matches("androidcoding") && getpwd.matches("androidcoding")){ 

       Intent i = new Intent(MainActivity.this,HomeScreen.class); 
       startActivity(i); 
      } else { 

       Toast.makeText(MainActivity.this, 
        "enter correct username & password", Toast.LENGTH_LONG).show(); 

      } 

     } 
     }); 

更多信息試試這個

loginButton = (ImageButton) findViewById(R.id.buttonLogin); 

Answer 2

喜http://androidcoding.in/2016/04/22/android-tutorial-on-login/