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Home/ Questions/Q 7978797
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T09:35:15+00:00 2026-06-04T09:35:15+00:00

I have 2 fragments, sharing the screen. In both, there is a layout that

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I have 2 fragments, sharing the screen.
In both, there is a layout that is listening to TouchEvents.
When I start a TouchEvent on the first fragment, and then moving my finger on top of the second one, the TouchEvent is still received by the first one, and never by the second.
I’ve tried different options without success. Could it be a solution, to handle the TouchEvent from the activity, and then dispatching it to both fragments ?
Any clue is welcome =)

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  1. Editorial Team
    Editorial Team
    2026-06-04T09:35:16+00:00Added an answer on June 4, 2026 at 9:35 am

    ok ! here is (very basically) how I managed that :
    MyActivity.java :

    private ArrayList<MyOnTouchListener> onTouchListeners = new  ArrayList<MusicShaker.MyOnTouchListener>(10);
    @Override
    public boolean dispatchTouchEvent(MotionEvent ev) {
        for (MyOnTouchListener listener : onTouchListeners) {
            listener.onTouch(ev);
        }
        return super.dispatchTouchEvent(ev);
    }
    public void registerMyOnTouchListener(MyOnTouchListener listener){
        onTouchListeners.add(listener);
    }
    public interface MyOnTouchListener {
        public void onTouch(MotionEvent ev);
    }
    

    Fragment1.java :

    ((MyActivity)getActivity()).registerMyOnTouchListener(new MyActivity.MyOnTouchListener() {
    
            @Override
            public void onTouch(MotionEvent ev) {
                Log.d("", "Fragment1 got it");
            }
        });
    

    Fragment2.java :

    ((MyActivity)getActivity()).registerMyOnTouchListener(new MyActivity.MyOnTouchListener() {
    
            @Override
            public void onTouch(MotionEvent ev) {
                Log.d("", "Fragment2 got it");
            }
        });
    

    It works, but if anyone has a better solution, I’m always listening =)

    Edit: be careful using this solution, as the MotionEvent x and y will be the same as rawX and rawY (no longer converted to View’s coordinates)

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