In my app, 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 isn't occurring 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 gives me 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.
check if you already added the view
The error message says what You should do.
The code below solved it for me:
Note: The error was from my fragment class and by overriding the onDestroy method like this, I could solve it.
I found another fix:
Here i just have an if statement check if the view had a parent and if it didn't Create the new dialog, set the contentView and show the dialog in my "createAlgorithmDialog()" method.
This also sets the positive and negative buttons (ok and cancel buttons) up with onClickListeners.
I came here on searching the error with my recyclerview but the solution didn't work (obviously). I have written the cause and the solution for it in case of recyclerview. Hope it helps someone.
The error is caused if in the
onCreateViewHolder()
the following method is followed:Instead it should be