单击图像上的自定义弹出窗口

时间:2014-02-17 07:21:50

标签: android popup

我想打开一个自定义弹出窗口,当我点击一个图像时,弹出窗口应该以给定的高度和宽度打开,我已经完成了编码完美但它对我来说不起作用任何人都可以帮我解决这个问题问题

这是我的代码,

ivmainmenu.setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View v) {
            // TODO Auto-generated method stub

            if (p != null)
                   showPopup(Home.this, p);
                 }

        private void showPopup(Home context, Point p) {
            // TODO Auto-generated method stub
            int popupWidth = 200;
               int popupHeight = 150;

               // Inflate the popup_layout.xml
               LinearLayout viewGroup = (LinearLayout) context.findViewById(R.id.popup_element);
               LayoutInflater layoutInflater = (LayoutInflater) context
                 .getSystemService(Context.LAYOUT_INFLATER_SERVICE);
               View layout = layoutInflater.inflate(R.layout.popupwindow, viewGroup);

               final PopupWindow popup = new PopupWindow(context);
               popup.setContentView(layout);
               popup.setWidth(popupWidth);
               popup.setHeight(popupHeight);
               popup.setFocusable(true);

               int OFFSET_X = 30;
               int OFFSET_Y = 30;

               popup.setBackgroundDrawable(new BitmapDrawable());

               popup.showAtLocation(layout, Gravity.NO_GRAVITY, p.x + OFFSET_X, p.y + OFFSET_Y);


        }
               });

0 个答案:

没有答案