Fix Python – Request UAC elevation from within a Python script?


Asked By – jwfearn

I want my Python script to copy files on Vista. When I run it from a normal cmd.exe window, no errors are generated, yet the files are NOT copied. If I run cmd.exe “as administator” and then run my script, it works fine.

This makes sense since User Account Control (UAC) normally prevents many file system actions.

Is there a way I can, from within a Python script, invoke a UAC elevation request (those dialogs that say something like “such and such app needs admin access, is this OK?”)

If that’s not possible, is there a way my script can at least detect that it is not elevated so it can fail gracefully?

Now we will see solution for issue: Request UAC elevation from within a Python script?


As of 2017, an easy method to achieve this is the following:

import ctypes, sys

def is_admin():
        return ctypes.windll.shell32.IsUserAnAdmin()
        return False

if is_admin():
    # Code of your program here
    # Re-run the program with admin rights
    ctypes.windll.shell32.ShellExecuteW(None, "runas", sys.executable, " ".join(sys.argv), None, 1)

If you are using Python 2.x, then you should replace the last line for:

ctypes.windll.shell32.ShellExecuteW(None, u"runas", unicode(sys.executable), unicode(" ".join(sys.argv)), None, 1)

Also note that if you converted you python script into an executable file (using tools like py2exe, cx_freeze, pyinstaller) then you should use sys.argv[1:] instead of sys.argv in the fourth parameter.

Some of the advantages here are:

  • No external libraries required. It only uses ctypes and sys from standard library.
  • Works on both Python 2 and Python 3.
  • There is no need to modify the file resources nor creating a manifest file.
  • If you don’t add code below if/else statement, the code won’t ever be executed twice.
  • You can get the return value of the API call in the last line and take an action if it fails (code <= 32). Check possible return values here.
  • You can change the display method of the spawned process modifying the sixth parameter.

Documentation for the underlying ShellExecute call is here.

This question is answered By – Martín De la Fuente

This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0