问如何使用桌面对象与更改窗口标题的应用程序连接

EN

import pywinauto
import subprocess
from pywinauto import Desktop, application
from pywinauto.application import Application

sp = subprocess.call("start shell:AppsFolder\SpotifyAB.SpotifyMusic_zpdnekdrzrea0!Spotify",shell=True)
app = pywinauto.Application(backend='uia').connect(path="Spotify.exe")

#this works only if there is NO song playing on Spotify; if any song is playing, 
#it fails to work since it can't find any app window with the title "Spotify"
#and the name of the window is instead the current song playing
#which I don't know what it is going to be at any given point
dlg = Desktop(backend="uia").Spotify
#does not work
dlg = Desktop(backend="uia").window(process=app.process)

dlg.child_window(title="Your Library", control_type="Hyperlink",found_index=0).wait("exists",30).invoke()

dlg.child_window(title="Orchestra", control_type="Hyperlink",found_index=0).wait("exists",30).invoke()
try:
    dlg.child_window(title="Play Orchestra", control_type="Button").wait("exists",5).invoke()
except:
    pass

    # this also doesn't work most of the time
    dlg.wait("",10).close()