如何获得Xlib.display.Window按id实例？

Question

我找到了以下代码（http://pastebin.com/rNkUj5V8），但我宁愿使用直接查找：

import Xlib
import Xlib.display

def get_window_by_id(winid):
    mydisplay = Xlib.display.Display()
    root = mydisplay.screen().root # should loop over all screens
    inspection_list = [root]

    while len(inspection_list) != 0:
        awin = inspection_list.pop(0)
        if awin.id == winid:
            return awin
        children = awin.query_tree().children
        if children != None:
            inspection_list += children

    return None

# use xwininfo -tree to click on something (panel was good for me)
# until you find a window with a name, then put that id in here
print get_window_by_id(0x1400003)
print get_window_by_id(0x1400003).get_wm_name()

我尝试过直接实例化一个窗口对象，但是调用get_attributes失败：

^{pr2}$

Answer 1

使用dpy.create_resource_object('window', 0x1400003)，其中dpy是一个Display对象，为具有给定XID的现有窗口在该显示器上获取一个Window对象。在

用法示例：

>>> import Xlib
>>> import Xlib.display
>>> dpy = Xlib.display.Display()
>>> win = dpy.create_resource_object('window', 0x277075e)
>>> win.get_wm_class()
('gnome-terminal', 'Gnome-terminal')