Java Android ListView在精确位置单击显示弹出窗口
我有一个Java Android ListView在精确位置单击显示弹出窗口,java,dynamic,android-listview,position,popupwindow,Java,Dynamic,Android Listview,Position,Popupwindow,我有一个弹出窗口,当我在列表视图上单击项目时,它显示为一个下拉列表: @Override public void onItemClick(AdapterView<?> parent, View view, int position, long id) { View v = getLayoutInflater().inflate(R.layout.popup_click_menu, null); final PopupWi
弹出窗口
,当我在列表视图上单击项目
时,它显示为一个下拉列表:
@Override
public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
View v = getLayoutInflater().inflate(R.layout.popup_click_menu, null);
final PopupWindow mypopupWindow = new PopupWindow(v,300, RelativeLayout.LayoutParams.WRAP_CONTENT, true);
mypopupWindow.showAsDropDown(view, -153, 0);
}
@覆盖
public void onItemClick(AdapterView父对象、视图、整型位置、长id){
视图v=GetLayoutFlater()。充气(R.layout.popup\u单击菜单,空);
final PopupWindow mypopupWindow=新的PopupWindow(v,300,RelativeLayout.LayoutParams.WRAP_内容,true);
mypopupWindow.showAsDropDown(视图,-153,0);
}
结果是:
我的问题是:我想有一个我的弹出窗口的动态位置。因此,如果单击“代码>项目< /代码>的中间,则应该显示在中间。我该怎么做呢?你可以用触摸监听器解决这个问题:
list.setOnTouchListener(new OnTouchListener() {
public boolean onTouch(View view, MotionEvent event) {
positionX = (int) event.getX();
return false; // not consumed; forward to onClick
}
});
然后您就有了确切的x位置,您可以通过onItemClick
移动PopupWindow
:
mypopupWindow.showAsDropDown(view, positionX, 0);