Question

我有一个按钮：＆＃34; btnOpenPopup＆＃34; （在滚动视图中），当我点击按钮时，会出现一个弹出窗口，我可以在屏幕上移动这个弹出窗口

这是我的代码：

final ImageButton btnOpenPopup = (ImageButton) findViewById(R.id.buttonUndoRedo);
    btnOpenPopup.setOnClickListener(new ImageButton.OnClickListener() {

        @Override
        public void onClick(View arg0) {

            if (popupWindowEnabled == true) {
                popupWindowEnabled = false;
                LayoutInflater layoutInflater =
                        (LayoutInflater) getBaseContext()
                                .getSystemService(LAYOUT_INFLATER_SERVICE);
                final View popupView = layoutInflater.inflate(R.layout.popup_window, null);
                final PopupWindow popupWindow = new PopupWindow(
                        popupView, ViewGroup.LayoutParams.WRAP_CONTENT, ViewGroup.LayoutParams.WRAP_CONTENT);

                // Font
                TextView txt = (TextView) popupView.findViewById(R.id.textViewPopupWindow);
                Typeface myNewFace = Typeface.createFromAsset(getAssets(), "fonts/futura.ttf");
                txt.setTypeface(myNewFace);

                Button btnDismiss = (Button) popupView.findViewById(R.id.dismiss);

                btnDismiss.setOnClickListener(new Button.OnClickListener() {
                    @Override
                    public void onClick(View v) {
                        popupWindow.dismiss();
                        popupWindowEnabled = true;
                    }
                });

                popupWindow.showAsDropDown(btnOpenPopup, 50, -30);

                popupView.setOnTouchListener(new View.OnTouchListener() {
                    int orgX, orgY;
                    int offsetX, offsetY;

                    @Override
                    public boolean onTouch(View v, MotionEvent event) {
                        //Toast.makeText(getApplicationContext(), "dEMO", Toast.LENGTH_SHORT).show();
                        switch (event.getAction()) {
                            case MotionEvent.ACTION_DOWN:
                                orgX = (int) event.getX();
                                orgY = (int) event.getY();
                                break;
                            case MotionEvent.ACTION_MOVE:
                                offsetX = (int) event.getRawX() - orgX;
                                offsetY = (int) event.getRawY() - orgY;
                                popupWindow.update(offsetX, offsetY, -1, -1, true);
                                break;
                        }
                        return true;
                    }
                });
            }

        }

    });

问题是当我在scrollview中移动btnOpenPopup时，弹出窗口会离开它的位置并跟随btnOpenPopup

当我们移动btnOpenPopup时，如何防止弹出窗口移动？提前谢谢！

Answer 1

我认为您需要更改.showAtLocation(..)参数。

更改

popupWindow.showAsDropDown(btnOpenPopup, 50, -30);

到

popupWindow.showAsDropDown(rootViewId, 50, -30);// RootView id is your main layout id.

我认为您需要添加 ScrollViewId 。

我希望它会帮助你。

弹出窗口Android

1 个答案: