通过使用Python在管理模式下打开应用程序实现自动化

要求： 使用脚本在管理模式下打开应用程序并自动执行其活动。 活动包括打开文件菜单->加载文件并执行所需任务

关于应用程序： 我试图自动化的应用程序是用C++ java、WPF和C++库开发的。 用于自动化的语言：Python

版本：2.7.14

使用的模块：Pywinauto、Pyautogui

尝试完成：

在应用程序的兼容性方面，它被设置为以管理员身份运行，并尝试使用以下代码打开应用程序，但导致了一组运行时错误

先决条件：

使用的代码：

from pywinauto.application import Application
import time
app = Application(backend="uia").start("C:\Program Files\**sample**.exe")
#app = Application(backend="uia").connect("C:\Program Files\**sample**.exe")
dlg = app.Update
time.sleep(2)
app.windows()
window = app.top_window()
print app.windows()
print('printing the control identifiers')
window.print_control_identifiers()

错误： 引发AppStartError（消息） pywinauto.application.AppStartError:无法创建进程“C:\Program Files*sample*.exe” CreateProcess返回错误：（740，“CreateProcess”，“请求的操作需要提升”。）

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尝试使用以下引用在管理模式下打开应用程序： 但我不能按要求实现自动化

结果：我能够以管理员权限打开应用程序，但pywinauto和pyautogui脚本失败

错误：无法将Pyautogui和pywinauto脚本与runAsAdmin（）一起使用

注意：为了打开应用程序并进行自动化，使用Pywinauto并尝试与应用程序菜单交互，失败了，因此制作了一个Pyautogui获取坐标的脚本

代码：

import sys, os, traceback, types
import pywinauto.controls.uia_controls
from pywinauto import mouse
import time
import os
import pyautogui
import pywinauto

def isUserAdmin():

    if os.name == 'nt':
        import ctypes
        # WARNING: requires Windows XP SP2 or higher!
        try:
            return ctypes.windll.shell32.IsUserAnAdmin()
        except:
            traceback.print_exc()
            print "Admin check failed, assuming not an admin."
            return False
    elif os.name == 'posix':
        # Check for root on Posix
        return os.getuid() == 0
    else:
        raise RuntimeError, "Unsupported operating system for this module: %s" % (os.name,)

def runAsAdmin(cmdLine=None, wait=True):

    if os.name != 'nt':
        raise RuntimeError, "This function is only implemented on Windows."

    import win32api, win32con, win32event, win32process
    from win32com.shell.shell import ShellExecuteEx
    from win32com.shell import shellcon

    python_exe = sys.executable

    if cmdLine is None:
        cmdLine = [python_exe] + sys.argv
    elif type(cmdLine) not in (types.TupleType,types.ListType):
        raise ValueError, "cmdLine is not a sequence."
    cmd = '"%s"' % (cmdLine[0],)
    # XXX TODO: isn't there a function or something we can call to massage command line params?
    params = " ".join(['"%s"' % (x,) for x in cmdLine[1:]])
    cmdDir = ''
    showCmd = win32con.SW_SHOWNORMAL
    #showCmd = win32con.SW_HIDE
    lpVerb = 'runas'  # causes UAC elevation prompt.

    # print "Running", cmd, params

    # ShellExecute() doesn't seem to allow us to fetch the PID or handle
    # of the process, so we can't get anything useful from it. Therefore
    # the more complex ShellExecuteEx() must be used.

    # procHandle = win32api.ShellExecute(0, lpVerb, cmd, params, cmdDir, showCmd)

    procInfo = ShellExecuteEx(nShow=showCmd,
                              fMask=shellcon.SEE_MASK_NOCLOSEPROCESS,
                              lpVerb=lpVerb,
                              lpFile=cmd,
                              lpParameters=params)

    if wait:
        procHandle = procInfo['hProcess']
        obj = win32event.WaitForSingleObject(procHandle, win32event.INFINITE)
        rc = win32process.GetExitCodeProcess(procHandle)
        #print "Process handle %s returned code %s" % (procHandle, rc)
    else:
        rc = None

    return rc

def test():
    rc = 0
    if not isUserAdmin():
        print "You're not an admin.", os.getpid(), "params: ", sys.argv
        rc = runAsAdmin(["C:\Program Files\**sample**.exe"])
        time.sleep(10)
        pywinauto.mouse.click(button='left', coords=(1199, 478))
        app1 = rc['sample']
        pywinauto.mouse.click(button='left', coords=(42, 33))
        time.sleep(10)
        print(pyautogui.position())

        rc = runAsAdmin()
    else:
        print "You are an admin!", os.getpid(), "params: ", sys.argv
        rc = 0
    x = raw_input('Press Enter to exit.')
    return rc


if __name__ == "__main__":
    sys.exit(test())

import pywinauto
from pywinauto import application, timings
import pyautogui
from pywinauto.keyboard import send_keys
from pywinauto.application import Application
import pywinauto.controls.uia_controls
from pywinauto import mouse
import win32api
import git
import time
import os
app = Application(backend="uia").start("C:\Program Files\**sample**.exe")
#app = Application(backend="uia").connect("C:\Program Files\**sample**.exe")
dlg = app.Update
time.sleep(2)
app.windows()
window = app.top_window()
print app.windows()
print('printing the control identifiers')
window.print_control_identifiers()
app1 = app[u'**sample**']
app2 = Application.connect(title=u'**sample**')
app1.print_control_identifiers()
apps = app1[u'File']
#print apps
dlg_spec = app.Untitled**sample**
print('dlg_spec :')
#print dlg_spec
print('Done sample')
#time.sleep(2)

3.尝试在没有管理员权限的情况下打开应用程序尝试使用自动化Pywinauto并尝试与应用程序菜单交互，但失败

代码：

import sys, os, traceback, types
import pywinauto.controls.uia_controls
from pywinauto import mouse
import time
import os
import pyautogui
import pywinauto

def isUserAdmin():

    if os.name == 'nt':
        import ctypes
        # WARNING: requires Windows XP SP2 or higher!
        try:
            return ctypes.windll.shell32.IsUserAnAdmin()
        except:
            traceback.print_exc()
            print "Admin check failed, assuming not an admin."
            return False
    elif os.name == 'posix':
        # Check for root on Posix
        return os.getuid() == 0
    else:
        raise RuntimeError, "Unsupported operating system for this module: %s" % (os.name,)

def runAsAdmin(cmdLine=None, wait=True):

    if os.name != 'nt':
        raise RuntimeError, "This function is only implemented on Windows."

    import win32api, win32con, win32event, win32process
    from win32com.shell.shell import ShellExecuteEx
    from win32com.shell import shellcon

    python_exe = sys.executable

    if cmdLine is None:
        cmdLine = [python_exe] + sys.argv
    elif type(cmdLine) not in (types.TupleType,types.ListType):
        raise ValueError, "cmdLine is not a sequence."
    cmd = '"%s"' % (cmdLine[0],)
    # XXX TODO: isn't there a function or something we can call to massage command line params?
    params = " ".join(['"%s"' % (x,) for x in cmdLine[1:]])
    cmdDir = ''
    showCmd = win32con.SW_SHOWNORMAL
    #showCmd = win32con.SW_HIDE
    lpVerb = 'runas'  # causes UAC elevation prompt.

    # print "Running", cmd, params

    # ShellExecute() doesn't seem to allow us to fetch the PID or handle
    # of the process, so we can't get anything useful from it. Therefore
    # the more complex ShellExecuteEx() must be used.

    # procHandle = win32api.ShellExecute(0, lpVerb, cmd, params, cmdDir, showCmd)

    procInfo = ShellExecuteEx(nShow=showCmd,
                              fMask=shellcon.SEE_MASK_NOCLOSEPROCESS,
                              lpVerb=lpVerb,
                              lpFile=cmd,
                              lpParameters=params)

    if wait:
        procHandle = procInfo['hProcess']
        obj = win32event.WaitForSingleObject(procHandle, win32event.INFINITE)
        rc = win32process.GetExitCodeProcess(procHandle)
        #print "Process handle %s returned code %s" % (procHandle, rc)
    else:
        rc = None

    return rc

def test():
    rc = 0
    if not isUserAdmin():
        print "You're not an admin.", os.getpid(), "params: ", sys.argv
        rc = runAsAdmin(["C:\Program Files\**sample**.exe"])
        time.sleep(10)
        pywinauto.mouse.click(button='left', coords=(1199, 478))
        app1 = rc['sample']
        pywinauto.mouse.click(button='left', coords=(42, 33))
        time.sleep(10)
        print(pyautogui.position())

        rc = runAsAdmin()
    else:
        print "You are an admin!", os.getpid(), "params: ", sys.argv
        rc = 0
    x = raw_input('Press Enter to exit.')
    return rc


if __name__ == "__main__":
    sys.exit(test())

import pywinauto
from pywinauto import application, timings
import pyautogui
from pywinauto.keyboard import send_keys
from pywinauto.application import Application
import pywinauto.controls.uia_controls
from pywinauto import mouse
import win32api
import git
import time
import os
app = Application(backend="uia").start("C:\Program Files\**sample**.exe")
#app = Application(backend="uia").connect("C:\Program Files\**sample**.exe")
dlg = app.Update
time.sleep(2)
app.windows()
window = app.top_window()
print app.windows()
print('printing the control identifiers')
window.print_control_identifiers()
app1 = app[u'**sample**']
app2 = Application.connect(title=u'**sample**')
app1.print_control_identifiers()
apps = app1[u'File']
#print apps
dlg_spec = app.Untitled**sample**
print('dlg_spec :')
#print dlg_spec
print('Done sample')
#time.sleep(2)

如果有人能使用Pywinauto来帮助脚本实现自动化，包括在管理模式下打开应用程序并与其中的菜单交互，那就太好了

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提前感谢

您必须在操作系统设置中手动禁用UAC。UAC确认对话框不能由操作系统设计自动执行（当然！）。禁用UAC后，您将看不到确认对话框，您的帖子中的配方（2）将起作用

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但是请注意，最好以admin的身份重新启动脚本，而不仅仅是以admin的身份启动应用程序。因为否则pywinauto非特权进程将无法访问应用程序的管理进程。

嗨，vasily ryabov，感谢您的回复。关于管理模式，我将尝试您的建议并返回。我已尝试使用Pyuac脚本打开应用程序：，效果很好。但我无法使用Pywinauto的应用程序模块继续进行菜单选择。