The specified child already has a parent. You must call removeView() on the child's parent first (Android)

blackst0ne picture blackst0ne · Jan 21, 2015 · Viewed 186k times · Source

I have to switch between two layouts frequently. The error is happening in the layout posted below.

When my layout is called the first time, there doesn't occur any error and everything's fine. When I then call a different layout (a blank one) and afterwards call my layout a second time, it throws the following error:

> FATAL EXCEPTION: main
>     java.lang.IllegalStateException: The specified child already has a parent. You must call removeView() on the child's parent first.

My layout-code looks like this:

    tv = new TextView(getApplicationContext()); // are initialized somewhere else
    et = new EditText(getApplicationContext()); // in the code


private void ConsoleWindow(){
        runOnUiThread(new Runnable(){

     @Override
     public void run(){

        // MY LAYOUT:
        setContentView(R.layout.activity_console);
        // LINEAR LAYOUT
        LinearLayout layout=new LinearLayout(getApplicationContext());
        layout.setOrientation(LinearLayout.VERTICAL);
        setContentView(layout);

        // TEXTVIEW
        layout.addView(tv); //  <==========  ERROR IN THIS LINE DURING 2ND RUN
        // EDITTEXT
        et.setHint("Enter Command");
        layout.addView(et);
        }
    }
}

I know this question has been asked before, but it didn't help in my case.

Answer

Zielony picture Zielony · Jan 21, 2015

The error message says what You should do.

// TEXTVIEW
if(tv.getParent() != null) {
    ((ViewGroup)tv.getParent()).removeView(tv); // <- fix
}
layout.addView(tv); //  <==========  ERROR IN THIS LINE DURING 2ND RUN
// EDITTEXT