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

0 votes

I am being required to constantly switch between two layouts frequently and despite that, the error is occurring in the layout mentioned below and also when my layout is called the ‘First Time’, then no errors occur and everything works fine but however, when I call a blank layout and later name it my ‘Second Time’, I receive the ERROR mentioned below:- 

> 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 me in this case.

Feb 16, 2022 in Java by Soham
• 9,700 points
4,110 views

1 answer to this question.

0 votes

The error message says what You should do. Use the following lines of code and the ERROR won't occur:

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

answered Feb 16, 2022 by Aditya
• 7,660 points

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