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.
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