python-如何在Windows上以提升的特权运行脚本？

我一直在试图弄清楚如何运行一堆都需要提升权限的应用程序.诸如DameWare,MSC.exe,PowerShell.exe和SCCM Manager控制台之类的应用程序都用于我的日常工作中.我现在正在运行Win7,并计划最终迁移到Win10.每天我运行这些...

我一直在试图弄清楚如何运行一堆都需要提升权限的应用程序.诸如DameWare,MSC.exe,PowerShell.exe和SCCM Manager控制台之类的应用程序都用于我的日常工作中.

我现在正在运行Win7,并计划最终迁移到Win10.每天我运行这些程序,一个一个地运行它们并为每个键入名称/密码是很费时间的.我想我只是“自动化无聊的东西”,然后让Python来做.

在此问题(How to run python script with elevated privilege on windows)上,答案就在那里,并且发布了名为“ admin”的旧模块的代码.但是,它是用Python 2编写的,在Python 3.5中不能很好地工作.我已经完成了我有限的python知识相关的工作,但是在尝试运行时却不断出错

Traceback (most recent call last):
  File "<pyshell#1>", line 1, in <module>
    runAsAdmin('cmd.exe')
  File "I:\Scripting\Python\admin.py", line 41, in runAsAdmin
    elif type(cmdLine) not in (types.TupleType,types.ListType):
AttributeError: module 'types' has no attribute 'TupleType'

我已经进行了一些研究,但我所能找到的只是Python 2文档或示例,而不是Python 3转换/等效文件.

这是admin.py的源代码,我已经尽力将其升级到Python 3.5了.您可以提供的任何帮助将不胜感激！

#!/usr/bin/env python
# -*- coding: utf-8; mode: python; py-indent-offset: 4; indent-tabs-mode: nil -*-
# vim: fileencoding=utf-8 tabstop=4 expandtab shiftwidth=4

# (C) COPYRIGHT ? Preston Landers 2010
# Released under the same license as Python 2.6.5


import sys, os, traceback, types

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


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

解决方法:

它看起来像type.TupleType和types.ListType在Python 3中不存在.请尝试以下操作：

elif type(cmdLine)不在(元组,列表)

说“ cmdLine不是序列”之后的值错误并不完全准确,因为字符串是序列,但实际上应该引发ValueError.我可能将其改写为“ cmdLine应该是一个非空的元组或列表,或者是None”.您可以对其进行更新,以更广泛地检查cmdLine是否为不可迭代的非字符串,但这可能会显得过大.