Android Dialog InvocationTargetException

Question

我想要一个对话框，用户可以输入用户名和密码来注册他的应用程序 对话框显示但如果我输入任何数据并单击我的确认按钮，异常即将出现=（ 有谁知道为什么我得到这个例外？

    Context mContext = getApplicationContext();
    LayoutInflater inflater = (LayoutInflater) mContext.getSystemService(LAYOUT_INFLATER_SERVICE);
    View layout = inflater.inflate(R.layout.dialog_register_app, (ViewGroup) findViewById(R.id.layout_root));

    //Start building dialog
    AlertDialog.Builder builder = new AlertDialog.Builder(this);
    builder.setTitle(getString(R.string.dialog_register_app_title));
    builder.setView(layout);
    builder.setCancelable(false);
    builder.setPositiveButton(getString(R.string.dialog_register_confirm_button_text), new DialogInterface.OnClickListener() {
        public void onClick(DialogInterface dialog, int id) {
            //Auslesen der EditText
            EditText oUsername = (EditText) findViewById(R.id.edit_username);
            String strUsername = oUsername.getText().toString(); // This is line where exception is calling
            //Do something with strUsername
        }
   });
   AlertDialog alert = builder.create();
   alert.show();
...
...
...

Answer 1

这应该可以使它工作，因为应用程序需要知道它需要从哪个布局（View）中选择带有id edit_username ....的edittext，因此应该在您创建的视图上调用findViewById方法，而不是查看当前活动...希望有所帮助

EditText oUsername = (EditText) layout.findViewById(R.id.edit_username);
            String strUsername = oUsername.getText().toString(); // This is line where exception is calling
...

Answer 2

如果此代码发生在Activity中，那么findViewById(...)可能无法按预期方式运行。您的活动布局中是否有EditText视图？

而是试试这个：

EditText oUsername = (EditText) layout.findViewById(R.id.edit_username);
String strUsername = oUsername.getText().toString();