我不能成爲一個消息，無論是成功地登錄與否

Question

EditText txtUserName; 
EditText txtPassword; 
Button btnLogin; 
Button btnCancel; 

    /** Called when the activity is first created. */ 
@Override 
public void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.main); 


    txtUserName=(EditText)this.findViewById(R.id.txtUname); 
    txtPassword=(EditText)this.findViewById(R.id.txtPwd); 
    btnLogin=(Button)this.findViewById(R.id.btnLogin); 

      Button btnLogin=(Button)this.findViewById(R.id.Button01); 
     btnLogin.setOnClickListener(new OnClickListener() { 

      @Override 
      public void onClick(View v) { 
       if((txtUserName.getText().toString()).equals(txtPassword.getText().toString())){ 
        Toast.makeText(LoginappActivity.this, "Login Successful",Toast.LENGTH_LONG).show(); 
        } else{ 
        Toast.makeText(LoginappActivity.this, "Invalid Login",Toast.LENGTH_LONG).show(); 
        } 
         } 
          }); 

      Button next = (Button) findViewById(R.id.Button01); 
      next.setOnClickListener(new View.OnClickListener() { 

       public void onClick(View view) { 
       Intent myIntent = new Intent(view.getContext(), AddName.class); 
       startActivityForResult(myIntent, 0); 
      }}); 

     } 

} 

，同時運行在模擬器這個應用程序運行良好，沒有任何錯誤，但它不顯示用戶名和密碼成功地登錄的任何消息或無效的登錄但是當我上單擊下一步按鈕，顯示下一個屏幕

Answer 1

刪除此行，並檢查它...

Button btnLogin=(Button)this.findViewById(R.id.Button01);<---------Remove it 

Answer 2

 Button btnLogin=(Button)this.findViewById(R.id.Button01); 

和

 Button next = (Button) findViewById(R.id.Button01); 

所以你在同一個按鈕上有兩個onClickListeners。