如何在AlertDialog输入期间防止我的应用崩溃

Question

我正在尝试在Android应用程序的开头创建2个Alertdialog窗口。我想要一个窗口询问用户名，第二个窗口显示用户名，但按提交按钮时我的应用程序仍然崩溃

当我想通过使用getText（）。toString（）;通过显示（Edit Text）的值来更改名为txt的TextView时，会发生主要问题。

public void Alert () {

    final AlertDialog.Builder builder = new AlertDialog.Builder(MainActivity.this, R.style.MyDialogTheme);

    View v = LayoutInflater.from(MainActivity.this).inflate(R.layout.dialog0, null);
    builder.setMessage("Welcome ! ");
    builder.setView(v);

    edt = (EditText) findViewById(R.id.edit_username);
    builder.setPositiveButton("Submit", new DialogInterface.OnClickListener() {

        @Override
        public void onClick(DialogInterface dialog, int which) {
Alert01();
        }
    })
            .setCancelable(false);
   AlertDialog alert = builder.create();

 alert.show();
}
public void Alert01 (){


        AlertDialog.Builder builder01 = new AlertDialog.Builder(MainActivity.this,R.style.MyDialogTheme);
        View v01 = LayoutInflater.from(MainActivity.this).inflate(R.layout.dialog1,null);
    builder01.setView(v01);

 txt.setText(getinfo());

        builder01.setTitle("Congratulations ! ");

        AlertDialog alert = builder01.create();
        alert.show();
    }

public String getinfo ()
{

    txt = (TextView) findViewById(R.id.text02);
    String info = edt.getText().toString();

    return info;
}
}