Android PopupWindow怎么合理控制弹出位置
1个回答
2016-09-23
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Android PopupWindow怎么合理控制弹出位置
private void showPopupWindow(View parent) {
if (popupWindow == null) {
LayoutInflater layoutInflater = (LayoutInflater) getSystemService(Context.LAYOUT_INFLATER_SERVICE);
view = layoutInflater.inflate(R.layout.group_list, null);
lv_group = (ListView) view.findViewById(R.id.lvGroup);
Collections.reverse(groups);
GroupAdapter groupAdapter = new GroupAdapter(this, groups);
lv_group.setAdapter(groupAdapter);
popupWindow = new PopupWindow(view, 200, 220);
}
popupWindow.setFocusable(true);
popupWindow.setOutsideTouchable(true); //设置点击屏幕其它地方弹出框消失
popupWindow.setBackgroundDrawable(new BitmapDrawable());
WindowManager windowManager = (WindowManager) getSystemService(Context.WINDOW_SERVICE);
int xPos = -popupWindow.getWidth() / 2
+ getCustomTitle().getCenter().getWidth() / 2;
popupWindow.showAsDropDown(parent, xPos, 4);
lv_group.setOnItemClickListener(new OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> adapterView, View view,
int position, long id) {
loadNew(((StringItem)(groups.get(position))).getId());
if (popupWindow != null)
popupWindow.dismiss();
}
});
}
只需要设置proupwindows的setOutsideTouchable属性即可。
以下为示例代码:
window.showAtLocation(parent, Gravity.RIGHT | Gravity.BOTTOM, 10,10);//显示位置
第一个参数指定PopupWindow的锚点view,即依附在哪个view上。
第二个参数指定起始点
第三个参数设置以起始点的右下角为原点,向左、上各偏移的像素。
自己看下API
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