2012-06-24 104 views
1

我刚刚定义了一个对话框作为android网站的建议,但它不显示按钮单击,一些代码scracht(我没有错误): static final int DIALOG_A = 1; static final int DIALOG_B = 2; TextView textX;showDialog不显示自定义对话框

protected Dialog onCreateDialog(int id, String text) { 
    Dialog dialog = new Dialog(this); 
    dialog.setContentView(decrytedText); 
    Log.v("DialogTest", " onCreateDialog(): +++ START +++"); 

    switch(id) 
    { 
     case DIALOG_A: 
      dialog.setTitle(this.getString(R.string.dialog_title_wrong_key)); 
      break; 
     case DIALOG_B: 
      dialog.setTitle(this.getString(R.string.dialog_title_ok_key)); 
      break; 
    } 
    return dialog; 
} 

somebutton.setOnClickListener(new View.OnClickListener() 
      { 
       public void onClick(View v) 
       { 
        Object butts[] = (Object []) v.getTag(); 

        try 
        { 
         Log.v("DialogTest", " try"); 
         String decrypted ="OK GO"; 
         textX.setText(decrypted); 
         showDialog(DIALOG_A); 
        } 
        catch (Exception e) 
        { 
         Log.v("DialogTest", "catch"); 
         textX.setText(R.string.dialog_no_wrong_key); 
         showDialog(DIALOG_B); 
        } 
       } 
      }); 
+0

编辑您的文章用的ShowDialog()方法的代码 –

+0

请告诉我常DIALOG_DECRYPT_TEXT指示? –

+0

我刚刚从另一个代码复制,我打算将其作为一般示例 – albanx

回答

3

我觉得应该有onCreateDialog的签名是这样的link

@Override 
protected Dialog onCreateDialog(int id) 

// or 

@Override 
protected Dialog onCreateDialog(int id, Bundle args) 

link1

+0

谢谢@Override忘记(但我不知道我必须放)实际上是问题 – albanx