Python 如何按id获取Xlib.display.Window实例？

我发现了以下代码（），但我更愿意使用直接查找：

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()

我已经尝试直接实例化窗口对象，但是调用

get\u attributes

失败：

w = Xlib.xobject.drawable.Window(Xlib.display.Display(), 67142278)
w.get_attributes()

/usr/lib/python2.7/dist-packages/Xlib/display.pyc in __getattr__(self, attr)
    211             return types.MethodType(function, self)
    212         except KeyError:
--> 213             raise AttributeError(attr)
    214 
    215     ###

AttributeError: send_request

---

使用

dpy。创建资源对象（'window'，0x1400003）

其中

dpy

是一个

Display

对象，以便为具有给定XID的现有窗口在该显示器上获取

window

对象

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')