how to dismiss a popupwindow when click button?

Question

• what i want to do

  i want to write a button that when i click it will show a popupwindow,and when i click the outside or click the button again the popupwindow will dismiss.

• what i do

  i use the code like this

  mPopupWindow=new PopupWindow();
  mPopupWindow.setTouchable(true);
  mPopupWindow.setOutsideTouchable(true);
  mPopupWindow.setWidth(WindowManager.LayoutParams.WRAP_CONTENT);
  mPopupWindow.setHeight(WindowManager.LayoutParams.WRAP_CONTENT);
  
      brush.setOnClickListener(new View.OnClickListener() {
      @Override
      public void onClick(View v) {
  
  
              mPopupWindow.setContentView(brushView);
              mPopupWindow.showAsDropDown(v);
  
      }
  });
  

• what the problem

  when i click the button again it will show Attempted to finish an input event but the input event receiver has already been disposed.

it think it because it tirgger the outside and clicklistener

Answer 1

To make the popup dissapear when clicked on the outside use:

   myPopupWindow.setOutsideTouchable(true);

To open and close the popup on clicking the button, put this code in your OnClickListener:

    if(myPopupWindow.isShowing()) {
        myPopupWindow.dismiss();
    } else {
        mPopupWindow.setContentView(brushView);
        mPopupWindow.showAsDropDown(v);
    }

Answer 2

just find a trick use

        mPopupWindow=new PopupWindow(new View(mcontext),WindowManager.LayoutParams.WRAP_CONTENT,WindowManager.LayoutParams.WRAP_CONTENT,true);

i don't know why but it works

Answer 3

Please try with below answer:

mPopupWindow.setBackgroundDrawable(new BitmapDrawable()); 
mPopupWindow.setOutsideTouchable(true); 

Answer 4

Just use

   myPopupWindow.setOutsideTouchable(true);