如何通过id获取Xlib.display.Window实例？

Question

我发现下面的代码（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() 

我曾尝试直接实例化一个Window对象，但随后调用get_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 

Answer 1

使用dpy.create_resource_object('window', 0x1400003)其中dpy是一个Display对象，该对象在显示器上为具有给定XID的现有窗口获取Window对象。

实例：

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