make a python program have my UID to modificate a file - python

So, in short, I created a script (named main.py) in which there was a moment where I wrote in a file. It worked well. However, the permissions on this file had to be rwxrwxrw- and hence anyone could modify the file on the server. That's not what I wanted. So I changed the permissions to rwxrwxr-- and then I changed the code of main.py :
#!/usr/bin/python
import subprocess
text = "I want this text to appear in my file"
command = subprocess.Popen(["python", "modificateFile.py", str('"')+text+str('"')], stderr=subprocess.PIPE, stdout=subprocess.PIPE) #I run another file that will do the task
for i in command.stderr:
print(i.decode("utf-8")) #check if there is any error
code of modificateFile.py
#!/usr/bin/python
import platform
import os
import sys
UID = 1080 #my UID.. I don't really know if it's the right way to program this
if __name__ == "__main__":
system = platform.system()
if system == "Linux": #ok it may be useless
os.setuid(UID) # /!\ I THINK THAT'S WHERE THE PROBLEM IS /!\
if len(sys.argv) > 1:
with open("file.txt", "w", encoding="utf-8") as f:
f.write(sys.argv[1])
else:
sys.stderr.write("not enough parameters to work") #didn't know which error I could raise..
#so as I already imported sys, I used this function
exit(-1)
else:
sys.stderr.write("wrong OS : program only work on linux")
exit(-1)
When I created this, I didn't know really what I was doing to be honest… I am learning programming.
Error message:
Traceback (most recent call last): File "modificateFile.py", line 11,
in os.setuid(UID) PermissionError: [Errno 1] Operation not permitted
I heard of SUID.. but I don't have root permissions.
Could someone explain what is wrong and what I could do ?
(if you need more elements, tell me it)

ok…. I thought SUID was only available for root programs… I had already tested it for this program but it didn't work. However, I learnt after that SUID is not available for programs with a shebang at the beginning… So, I created a code in C to do that.

Related

I keep having a error when running the script. It keeps sending "name input_file is not defined"

I am trying to create a python def to unzip a few .gz files within a folder. I know that the main script works if it is not in the def. The script I have created is similar to others I have done but this one give me the error
File "unzip.py", line 24, in
decompressed_files(input_folder)
NameError: name 'input_folder' is not defined
I copied the script below so someone can help me to see where the error is. I haven't done any BioInformatics for the last couple of years and I am a bit rusty.
import glob
import sys
import os
import argparse
import subprocess
import gzip
def decompressed_files(input_folder):
print ('starting decompressed_files')
output_folder=input_folder + '/fasta_files'
if os.path.exists(output_folder):
print ('folder already exists')
else:
os.makedirs(output_folder)
for f in input_folder:
fastqs=glob.glob(input_folder + '/*.fastq.gz')
cmd =[gunzip, -k, fastqs, output_folder]
my_file=subprocess.Popen(cmd)
my_file.wait
print ('The programme has finished doing its job')
decompressed_files(input_folder)
This is done for python 2.7, I know that is old but it is the one that it is installed in my work server.
That's why when you call decompressed_files(input_folder) in the last line, you didn't define input_folder before. you should do it like this :
input_folder = 'C:/Some Address/'
decompressed_files(input_folder)

FileNotFoundError When file exists (when created in current script)

I am trying to create a secure (e.g., SSL/HTTPS) XML-RPC Client Server. The client-server part works perfectly when the required certificates are present on my system; however, when I try to create the certificates during execution, I receive a FileNotFoundError when opening the ssl-wrapped socket even though the certificates are clearly present (because the preceding function created them.)
Why is the FileNotFoundError given when the files are present? (If I simply close and restart the python script no error is produced when opening the socket and everything works with no issue whatsoever.)
I've searched elsewhere for solutions, but the best/closest answer I've found is, perhaps, "race conditions" between creating the certificates and opening them. However, I've tried adding "sleep" to alleviate the possibility of race conditions (as well as running each function individually via a user input menu) with the same error every time.
What I am missing?
Here is a snippet of my code:
import os
import threading
import ssl
from xmlrpc.server import SimpleXMLRPCServer
import certs.gencert as gencert # <---- My python module for generating certs
...
rootDomain = "mydomain"
CERTFILE = "certs/mydomain.cert"
KEYFILE = "certs/mydomain.key"
...
def listenNow(ipAdd, portNum, serverCert, serverKey):
# Create XMLRPC Server, based on ipAdd/port received
server = SimpleXMLRPCServer((ipAdd, portNum))
# **THIS** is what causes the FileNotFoundError ONLY if
# the certificates are created during THE SAME execution
# of the program.
server.socket = ssl.wrap_socket(server.socket,
certfile=serverCert,
keyfile=serverKey,
do_handshake_on_connect=True,
server_side=True)
...
# Start server listening [forever]
server.serve_forever()
...
# Verify Certificates are present; if not present,
# create new certificates
def verifyCerts():
# If cert or key file not present, create new certs
if not os.path.isfile(CERTFILE) or not os.path.isfile(KEYFILE):
# NOTE: This [genert] will create certificates matching
# the file names listed in CERTFILE and KEYFILE at the top
gencert.gencert(rootDomain)
print("Certfile(s) NOT present; new certs created.")
else:
print("Certfiles Verified Present")
# Start a thread to run server connection as a daemon
def startServer(hostIP, serverPort):
# Verify certificates present prior to starting server
verifyCerts()
# Now, start thread
t = threading.Thread(name="ServerDaemon",
target=listenNow,
args=(hostIP,
serverPort,
CERTFILE,
KEYFILE
)
)
t.daemon = True
t.start()
if __name__ == '__main__':
startServer("127.0.0.1", 12345)
time.sleep(60) # <--To allow me to connect w/client before closing
When I run the above, with NO certificates present, this is the error I receive:
$ python3 test.py
Certfile(s) NOT present; new certs created.
Exception in thread ServerDaemon:
Traceback (most recent call last):
File "/usr/lib/python3.5/threading.py", line 914, in _bootstrap_inner
self.run()
File "/usr/lib/python3.5/threading.py", line 862, in run
self._target(*self._args, **self._kwargs)
File "test.py", line 41, in listenNow
server_side=True)
File "/usr/lib/python3.5/ssl.py", line 1069, in wrap_socket
ciphers=ciphers)
File "/usr/lib/python3.5/ssl.py", line 691, in __init__
self._context.load_cert_chain(certfile, keyfile)
FileNotFoundError: [Errno 2] No such file or directory
When I simply re-run the script a second time (i.e., the cert files are already present when it starts, everything runs as expected with NO errors, and I can connect my client just fine.
$ python3 test.py
Certfiles Verified Present
What is preventing the ssl.wrap_socket function from seeing/accessing the files that were just created (and thus producing the FileNotFoundError exception)?
EDIT 1:
Thanks for the comments John Gordon. Here is a copy of gencert.py, courtesy of Atul Varm, found here https://gist.github.com/toolness/3073310
import os
import sys
import hashlib
import subprocess
import datetime
OPENSSL_CONFIG_TEMPLATE = """
prompt = no
distinguished_name = req_distinguished_name
req_extensions = v3_req
[ req_distinguished_name ]
C = US
ST = IL
L = Chicago
O = Toolness
OU = Experimental Software Authority
CN = %(domain)s
emailAddress = varmaa#toolness.com
[ v3_req ]
# Extensions to add to a certificate request
basicConstraints = CA:FALSE
keyUsage = nonRepudiation, digitalSignature, keyEncipherment
subjectAltName = #alt_names
[ alt_names ]
DNS.1 = %(domain)s
DNS.2 = *.%(domain)s
"""
MYDIR = os.path.abspath(os.path.dirname(__file__))
OPENSSL = '/usr/bin/openssl'
KEY_SIZE = 1024
DAYS = 3650
CA_CERT = 'ca.cert'
CA_KEY = 'ca.key'
# Extra X509 args. Consider using e.g. ('-passin', 'pass:blah') if your
# CA password is 'blah'. For more information, see:
#
# http://www.openssl.org/docs/apps/openssl.html#PASS_PHRASE_ARGUMENTS
X509_EXTRA_ARGS = ()
def openssl(*args):
cmdline = [OPENSSL] + list(args)
subprocess.check_call(cmdline)
def gencert(domain, rootdir=MYDIR, keysize=KEY_SIZE, days=DAYS,
ca_cert=CA_CERT, ca_key=CA_KEY):
def dfile(ext):
return os.path.join('domains', '%s.%s' % (domain, ext))
os.chdir(rootdir)
if not os.path.exists('domains'):
os.mkdir('domains')
if not os.path.exists(dfile('key')):
openssl('genrsa', '-out', dfile('key'), str(keysize))
# EDIT 3: mydomain.key gets output here during execution
config = open(dfile('config'), 'w')
config.write(OPENSSL_CONFIG_TEMPLATE % {'domain': domain})
config.close()
# EDIT 3: mydomain.config gets output here during execution
openssl('req', '-new', '-key', dfile('key'), '-out', dfile('request'),
'-config', dfile('config'))
# EDIT 3: mydomain.request gets output here during execution
openssl('x509', '-req', '-days', str(days), '-in', dfile('request'),
'-CA', ca_cert, '-CAkey', ca_key,
'-set_serial',
'0x%s' % hashlib.md5(domain +
str(datetime.datetime.now())).hexdigest(),
'-out', dfile('cert'),
'-extensions', 'v3_req', '-extfile', dfile('config'),
*X509_EXTRA_ARGS)
# EDIT 3: mydomain.cert gets output here during execution
print "Done. The private key is at %s, the cert is at %s, and the " \
"CA cert is at %s." % (dfile('key'), dfile('cert'), ca_cert)
if __name__ == "__main__":
if len(sys.argv) < 2:
print "usage: %s <domain-name>" % sys.argv[0]
sys.exit(1)
gencert(sys.argv[1])
EDIT 2:
Regarding John's comment, "this might mean that those files are being created, but not in the directory [I] expect":
When I have the directory open in another window, I see the files pop up in the correct location during execution. In addition, when running the test.py script a second time with no changes, the files are identified as present in the correct (the same) location. This leads me to believe that file location is not the problem. Thanks for the suggestion. I'll keep looking.
EDIT 3:
I stepped through the gencert.py program, and each of the files are correctly output at the right time during execution. I indicated when exactly they were output within the above file, labeled with "EDIT 3"
While gencert is paused awaiting my input (raw_input), I can open/view/edit the mentioned files in another program with no problem.
In addition, with the first test.py instance running (paused, waiting for user input, just after mydomain.cert appears), I can run a second instance of test.py in another terminal and it sees/uses the files just fine.
Within the first instance, however, if I continue the program it outputs "FileNotFoundError."
The problem contained in the above stems from the use of os.chdir(rootdir) as suggested by John; however, the specifics are slightly different than the created files being in the wrong location. The problem is the current working directory (cwd) of the running program being changed by gencert(). Here are the specifics:
The program is started with test.py, which calls verifyCerts(). At this point the program is running in the current directory of whichever folder test.py is running inside of. Use os.getcwd() to find the current directory at this point. In this case (as an example), it is running in:
/home/name/testfolder/
Next, os.path.isfile() looks for the files "certs/mydomain.cert" and "certs/mydomain.key"; based on the file path above [e.g., the cwd], it is looking for the following files:
/home/name/testfolder/certs/mydomain.cert
/home/name/testfolder/certs/mydomain.key
The above files are not present, so the program executes gencert.gencert(rootDomain) which correctly creates both files as expected in the exact locations mentioned above in number 2.
The problem is indeed the os.chdir() call: When gencert() executes, it uses os.chdir() to change the cwd to "rootdir," which is os.path.abspath(os.path.dirname(__file__)), which is the directory of the current file (gencert.py). Since this file is located a folder deeper, the new cwd becomes:
/home/name/testfolder/certs/
When gencert() finishes execution and the control returns to test.py, the cwd never changes again; the cwd remains as /home/name/testfolder/certs/ even when execution returns to test.py.
Later, when ssl.wrap_socket() tries to find the serverCert and serverKey, it looks for "certs/mydomain.cert" and "certs/mydomain.key" in the cwd, so here is the full path of what it is looking for:
/home/name/testfolder/certs/certs/mydomain.cert
/home/name/testfolder/certs/certs/mydomain.key
These two files are NOT present, so the program correctly returns "FileNotFoundError".
Solution
A) Move the "gencert.py" file to the same directory as "test.py"
B) At the beginning of "gencert.py" add cwd = os.getcwd() to record the original cwd of the program; then, at the very end, add os.chdir(cwd) to change back to the original cwd before ending and giving control back to the calling program.
I went with option 'B', and my program now works flawlessly. I appreciate the assistance from John Gordon to point me toward finding the source of my problem.

Python: Run Script Under Same Window

I am trying to setup a program where when someone enters a command it will run that command which is a script in a sub folder called "lib".
Here is my code:
import os
while 1:
cmd = input(' >: ')
for file in os.listdir('lib'):
if file.endswith('.py'):
try:
os.system(str(cmd + '.py'))
except FileNotFoundError:
print('Command Not Found.')
I have a file: lib/new_user.py But when I try to run it I get this error:
Traceback (most recent call last):
File "C:/Users/Daniel/Desktop/Wasm/Exec.py", line 8, in <module>
exec(str(cmd + '.py'))
File "<string>", line 1, in <module>
NameError: name 'new_user' is not defined
Does anyone know a way around this? I would prefer if the script would be able to be executed under the same window so it doesn't open a completely new one up to run the code there. This may be a really Noob question but I have not been able to find anything on this.
Thanks,
Daniel Alexander
os.system(os.path.join('lib', cmd + '.py'))
You're invoking new_user.py but it is not in the current directory. You need to construct lib/new_user.py.
(I'm not sure what any of this has to do with windows.)
However, a better approach for executing Python code from Python is making them into modules and using import:
import importlib
cmd_module = importlib.import_module(cmd, 'lib')
cmd_module.execute()
(Assuming you have a function execute defined in lib/new_user.py)

Copying a file in python : Permission denied [duplicate]

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?
As of 2017, an easy method to achieve this is the following:
import ctypes, sys
def is_admin():
try:
return ctypes.windll.shell32.IsUserAnAdmin()
except:
return False
if is_admin():
# Code of your program here
else:
# 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.
It took me a little while to get dguaraglia's answer working, so in the interest of saving others time, here's what I did to implement this idea:
import os
import sys
import win32com.shell.shell as shell
ASADMIN = 'asadmin'
if sys.argv[-1] != ASADMIN:
script = os.path.abspath(sys.argv[0])
params = ' '.join([script] + sys.argv[1:] + [ASADMIN])
shell.ShellExecuteEx(lpVerb='runas', lpFile=sys.executable, lpParameters=params)
sys.exit(0)
It seems there's no way to elevate the application privileges for a while for you to perform a particular task. Windows needs to know at the start of the program whether the application requires certain privileges, and will ask the user to confirm when the application performs any tasks that need those privileges. There are two ways to do this:
Write a manifest file that tells Windows the application might require some privileges
Run the application with elevated privileges from inside another program
This two articles explain in much more detail how this works.
What I'd do, if you don't want to write a nasty ctypes wrapper for the CreateElevatedProcess API, is use the ShellExecuteEx trick explained in the Code Project article (Pywin32 comes with a wrapper for ShellExecute). How? Something like this:
When your program starts, it checks if it has Administrator privileges, if it doesn't it runs itself using the ShellExecute trick and exits immediately, if it does, it performs the task at hand.
As you describe your program as a "script", I suppose that's enough for your needs.
Cheers.
Just adding this answer in case others are directed here by Google Search as I was.
I used the elevate module in my Python script and the script executed with Administrator Privileges in Windows 10.
https://pypi.org/project/elevate/
The following example builds on MARTIN DE LA FUENTE SAAVEDRA's excellent work and accepted answer. In particular, two enumerations are introduced. The first allows for easy specification of how an elevated program is to be opened, and the second helps when errors need to be easily identified. Please note that if you want all command line arguments passed to the new process, sys.argv[0] should probably be replaced with a function call: subprocess.list2cmdline(sys.argv).
#! /usr/bin/env python3
import ctypes
import enum
import subprocess
import sys
# Reference:
# msdn.microsoft.com/en-us/library/windows/desktop/bb762153(v=vs.85).aspx
# noinspection SpellCheckingInspection
class SW(enum.IntEnum):
HIDE = 0
MAXIMIZE = 3
MINIMIZE = 6
RESTORE = 9
SHOW = 5
SHOWDEFAULT = 10
SHOWMAXIMIZED = 3
SHOWMINIMIZED = 2
SHOWMINNOACTIVE = 7
SHOWNA = 8
SHOWNOACTIVATE = 4
SHOWNORMAL = 1
class ERROR(enum.IntEnum):
ZERO = 0
FILE_NOT_FOUND = 2
PATH_NOT_FOUND = 3
BAD_FORMAT = 11
ACCESS_DENIED = 5
ASSOC_INCOMPLETE = 27
DDE_BUSY = 30
DDE_FAIL = 29
DDE_TIMEOUT = 28
DLL_NOT_FOUND = 32
NO_ASSOC = 31
OOM = 8
SHARE = 26
def bootstrap():
if ctypes.windll.shell32.IsUserAnAdmin():
main()
else:
# noinspection SpellCheckingInspection
hinstance = ctypes.windll.shell32.ShellExecuteW(
None,
'runas',
sys.executable,
subprocess.list2cmdline(sys.argv),
None,
SW.SHOWNORMAL
)
if hinstance <= 32:
raise RuntimeError(ERROR(hinstance))
def main():
# Your Code Here
print(input('Echo: '))
if __name__ == '__main__':
bootstrap()
Recognizing this question was asked years ago, I think a more elegant solution is offered on github by frmdstryr using his module pywinutils:
Excerpt:
import pythoncom
from win32com.shell import shell,shellcon
def copy(src,dst,flags=shellcon.FOF_NOCONFIRMATION):
""" Copy files using the built in Windows File copy dialog
Requires absolute paths. Does NOT create root destination folder if it doesn't exist.
Overwrites and is recursive by default
#see http://msdn.microsoft.com/en-us/library/bb775799(v=vs.85).aspx for flags available
"""
# #see IFileOperation
pfo = pythoncom.CoCreateInstance(shell.CLSID_FileOperation,None,pythoncom.CLSCTX_ALL,shell.IID_IFileOperation)
# Respond with Yes to All for any dialog
# #see http://msdn.microsoft.com/en-us/library/bb775799(v=vs.85).aspx
pfo.SetOperationFlags(flags)
# Set the destionation folder
dst = shell.SHCreateItemFromParsingName(dst,None,shell.IID_IShellItem)
if type(src) not in (tuple,list):
src = (src,)
for f in src:
item = shell.SHCreateItemFromParsingName(f,None,shell.IID_IShellItem)
pfo.CopyItem(item,dst) # Schedule an operation to be performed
# #see http://msdn.microsoft.com/en-us/library/bb775780(v=vs.85).aspx
success = pfo.PerformOperations()
# #see sdn.microsoft.com/en-us/library/bb775769(v=vs.85).aspx
aborted = pfo.GetAnyOperationsAborted()
return success is None and not aborted
This utilizes the COM interface and automatically indicates that admin privileges are needed with the familiar dialog prompt that you would see if you were copying into a directory where admin privileges are required and also provides the typical file progress dialog during the copy operation.
This may not completely answer your question but you could also try using the Elevate Command Powertoy in order to run the script with elevated UAC privileges.
http://technet.microsoft.com/en-us/magazine/2008.06.elevation.aspx
I think if you use it it would look like 'elevate python yourscript.py'
You can make a shortcut somewhere and as the target use:
python yourscript.py
then under properties and advanced select run as administrator.
When the user executes the shortcut it will ask them to elevate the application.
A variation on Jorenko's work above allows the elevated process to use the same console (but see my comment below):
def spawn_as_administrator():
""" Spawn ourself with administrator rights and wait for new process to exit
Make the new process use the same console as the old one.
Raise Exception() if we could not get a handle for the new re-run the process
Raise pywintypes.error() if we could not re-spawn
Return the exit code of the new process,
or return None if already running the second admin process. """
#pylint: disable=no-name-in-module,import-error
import win32event, win32api, win32process
import win32com.shell.shell as shell
if '--admin' in sys.argv:
return None
script = os.path.abspath(sys.argv[0])
params = ' '.join([script] + sys.argv[1:] + ['--admin'])
SEE_MASK_NO_CONSOLE = 0x00008000
SEE_MASK_NOCLOSE_PROCESS = 0x00000040
process = shell.ShellExecuteEx(lpVerb='runas', lpFile=sys.executable, lpParameters=params, fMask=SEE_MASK_NO_CONSOLE|SEE_MASK_NOCLOSE_PROCESS)
hProcess = process['hProcess']
if not hProcess:
raise Exception("Could not identify administrator process to install drivers")
# It is necessary to wait for the elevated process or else
# stdin lines are shared between 2 processes: they get one line each
INFINITE = -1
win32event.WaitForSingleObject(hProcess, INFINITE)
exitcode = win32process.GetExitCodeProcess(hProcess)
win32api.CloseHandle(hProcess)
return exitcode
This is mostly an upgrade to Jorenko's answer, that allows to use parameters with spaces in Windows, but should also work fairly well on Linux :)
Also, will work with cx_freeze or py2exe since we don't use __file__ but sys.argv[0] as executable
[EDIT]
Disclaimer: The code in this post is outdated.
I have published the elevation code as a python package.
Install with pip install command_runner
Usage:
from command_runner.elevate import elevate
def main():
"""My main function that should be elevated"""
print("Who's the administrator, now ?")
if __name__ == '__main__':
elevate(main)
[/EDIT]
import sys,ctypes,platform
def is_admin():
try:
return ctypes.windll.shell32.IsUserAnAdmin()
except:
raise False
if __name__ == '__main__':
if platform.system() == "Windows":
if is_admin():
main(sys.argv[1:])
else:
# Re-run the program with admin rights, don't use __file__ since py2exe won't know about it
# Use sys.argv[0] as script path and sys.argv[1:] as arguments, join them as lpstr, quoting each parameter or spaces will divide parameters
lpParameters = ""
# Litteraly quote all parameters which get unquoted when passed to python
for i, item in enumerate(sys.argv[0:]):
lpParameters += '"' + item + '" '
try:
ctypes.windll.shell32.ShellExecuteW(None, "runas", sys.executable, lpParameters , None, 1)
except:
sys.exit(1)
else:
main(sys.argv[1:])
For one-liners, put the code to where you need UAC.
Request UAC, if failed, keep running:
import ctypes, sys
ctypes.windll.shell32.IsUserAnAdmin() or ctypes.windll.shell32.ShellExecuteW(
None, "runas", sys.executable, " ".join(sys.argv), None, 1) > 32 and exit()
Request UAC, if failed, exit:
import ctypes, sys
ctypes.windll.shell32.IsUserAnAdmin() or (ctypes.windll.shell32.ShellExecuteW(
None, "runas", sys.executable, " ".join(sys.argv), None, 1) > 32, exit())
Function style:
# Created by BaiJiFeiLong#gmail.com at 2022/6/24
import ctypes
import sys
def request_uac_or_skip():
ctypes.windll.shell32.IsUserAnAdmin() or ctypes.windll.shell32.ShellExecuteW(
None, "runas", sys.executable, " ".join(sys.argv), None, 1) > 32 and sys.exit()
def request_uac_or_exit():
ctypes.windll.shell32.IsUserAnAdmin() or (ctypes.windll.shell32.ShellExecuteW(
None, "runas", sys.executable, " ".join(sys.argv), None, 1) > 32, sys.exit())
If your script always requires an Administrator's privileges then:
runas /user:Administrator "python your_script.py"

Request UAC elevation from within a Python script?

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?
As of 2017, an easy method to achieve this is the following:
import ctypes, sys
def is_admin():
try:
return ctypes.windll.shell32.IsUserAnAdmin()
except:
return False
if is_admin():
# Code of your program here
else:
# 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.
It took me a little while to get dguaraglia's answer working, so in the interest of saving others time, here's what I did to implement this idea:
import os
import sys
import win32com.shell.shell as shell
ASADMIN = 'asadmin'
if sys.argv[-1] != ASADMIN:
script = os.path.abspath(sys.argv[0])
params = ' '.join([script] + sys.argv[1:] + [ASADMIN])
shell.ShellExecuteEx(lpVerb='runas', lpFile=sys.executable, lpParameters=params)
sys.exit(0)
It seems there's no way to elevate the application privileges for a while for you to perform a particular task. Windows needs to know at the start of the program whether the application requires certain privileges, and will ask the user to confirm when the application performs any tasks that need those privileges. There are two ways to do this:
Write a manifest file that tells Windows the application might require some privileges
Run the application with elevated privileges from inside another program
This two articles explain in much more detail how this works.
What I'd do, if you don't want to write a nasty ctypes wrapper for the CreateElevatedProcess API, is use the ShellExecuteEx trick explained in the Code Project article (Pywin32 comes with a wrapper for ShellExecute). How? Something like this:
When your program starts, it checks if it has Administrator privileges, if it doesn't it runs itself using the ShellExecute trick and exits immediately, if it does, it performs the task at hand.
As you describe your program as a "script", I suppose that's enough for your needs.
Cheers.
Just adding this answer in case others are directed here by Google Search as I was.
I used the elevate module in my Python script and the script executed with Administrator Privileges in Windows 10.
https://pypi.org/project/elevate/
The following example builds on MARTIN DE LA FUENTE SAAVEDRA's excellent work and accepted answer. In particular, two enumerations are introduced. The first allows for easy specification of how an elevated program is to be opened, and the second helps when errors need to be easily identified. Please note that if you want all command line arguments passed to the new process, sys.argv[0] should probably be replaced with a function call: subprocess.list2cmdline(sys.argv).
#! /usr/bin/env python3
import ctypes
import enum
import subprocess
import sys
# Reference:
# msdn.microsoft.com/en-us/library/windows/desktop/bb762153(v=vs.85).aspx
# noinspection SpellCheckingInspection
class SW(enum.IntEnum):
HIDE = 0
MAXIMIZE = 3
MINIMIZE = 6
RESTORE = 9
SHOW = 5
SHOWDEFAULT = 10
SHOWMAXIMIZED = 3
SHOWMINIMIZED = 2
SHOWMINNOACTIVE = 7
SHOWNA = 8
SHOWNOACTIVATE = 4
SHOWNORMAL = 1
class ERROR(enum.IntEnum):
ZERO = 0
FILE_NOT_FOUND = 2
PATH_NOT_FOUND = 3
BAD_FORMAT = 11
ACCESS_DENIED = 5
ASSOC_INCOMPLETE = 27
DDE_BUSY = 30
DDE_FAIL = 29
DDE_TIMEOUT = 28
DLL_NOT_FOUND = 32
NO_ASSOC = 31
OOM = 8
SHARE = 26
def bootstrap():
if ctypes.windll.shell32.IsUserAnAdmin():
main()
else:
# noinspection SpellCheckingInspection
hinstance = ctypes.windll.shell32.ShellExecuteW(
None,
'runas',
sys.executable,
subprocess.list2cmdline(sys.argv),
None,
SW.SHOWNORMAL
)
if hinstance <= 32:
raise RuntimeError(ERROR(hinstance))
def main():
# Your Code Here
print(input('Echo: '))
if __name__ == '__main__':
bootstrap()
Recognizing this question was asked years ago, I think a more elegant solution is offered on github by frmdstryr using his module pywinutils:
Excerpt:
import pythoncom
from win32com.shell import shell,shellcon
def copy(src,dst,flags=shellcon.FOF_NOCONFIRMATION):
""" Copy files using the built in Windows File copy dialog
Requires absolute paths. Does NOT create root destination folder if it doesn't exist.
Overwrites and is recursive by default
#see http://msdn.microsoft.com/en-us/library/bb775799(v=vs.85).aspx for flags available
"""
# #see IFileOperation
pfo = pythoncom.CoCreateInstance(shell.CLSID_FileOperation,None,pythoncom.CLSCTX_ALL,shell.IID_IFileOperation)
# Respond with Yes to All for any dialog
# #see http://msdn.microsoft.com/en-us/library/bb775799(v=vs.85).aspx
pfo.SetOperationFlags(flags)
# Set the destionation folder
dst = shell.SHCreateItemFromParsingName(dst,None,shell.IID_IShellItem)
if type(src) not in (tuple,list):
src = (src,)
for f in src:
item = shell.SHCreateItemFromParsingName(f,None,shell.IID_IShellItem)
pfo.CopyItem(item,dst) # Schedule an operation to be performed
# #see http://msdn.microsoft.com/en-us/library/bb775780(v=vs.85).aspx
success = pfo.PerformOperations()
# #see sdn.microsoft.com/en-us/library/bb775769(v=vs.85).aspx
aborted = pfo.GetAnyOperationsAborted()
return success is None and not aborted
This utilizes the COM interface and automatically indicates that admin privileges are needed with the familiar dialog prompt that you would see if you were copying into a directory where admin privileges are required and also provides the typical file progress dialog during the copy operation.
This may not completely answer your question but you could also try using the Elevate Command Powertoy in order to run the script with elevated UAC privileges.
http://technet.microsoft.com/en-us/magazine/2008.06.elevation.aspx
I think if you use it it would look like 'elevate python yourscript.py'
You can make a shortcut somewhere and as the target use:
python yourscript.py
then under properties and advanced select run as administrator.
When the user executes the shortcut it will ask them to elevate the application.
A variation on Jorenko's work above allows the elevated process to use the same console (but see my comment below):
def spawn_as_administrator():
""" Spawn ourself with administrator rights and wait for new process to exit
Make the new process use the same console as the old one.
Raise Exception() if we could not get a handle for the new re-run the process
Raise pywintypes.error() if we could not re-spawn
Return the exit code of the new process,
or return None if already running the second admin process. """
#pylint: disable=no-name-in-module,import-error
import win32event, win32api, win32process
import win32com.shell.shell as shell
if '--admin' in sys.argv:
return None
script = os.path.abspath(sys.argv[0])
params = ' '.join([script] + sys.argv[1:] + ['--admin'])
SEE_MASK_NO_CONSOLE = 0x00008000
SEE_MASK_NOCLOSE_PROCESS = 0x00000040
process = shell.ShellExecuteEx(lpVerb='runas', lpFile=sys.executable, lpParameters=params, fMask=SEE_MASK_NO_CONSOLE|SEE_MASK_NOCLOSE_PROCESS)
hProcess = process['hProcess']
if not hProcess:
raise Exception("Could not identify administrator process to install drivers")
# It is necessary to wait for the elevated process or else
# stdin lines are shared between 2 processes: they get one line each
INFINITE = -1
win32event.WaitForSingleObject(hProcess, INFINITE)
exitcode = win32process.GetExitCodeProcess(hProcess)
win32api.CloseHandle(hProcess)
return exitcode
This is mostly an upgrade to Jorenko's answer, that allows to use parameters with spaces in Windows, but should also work fairly well on Linux :)
Also, will work with cx_freeze or py2exe since we don't use __file__ but sys.argv[0] as executable
[EDIT]
Disclaimer: The code in this post is outdated.
I have published the elevation code as a python package.
Install with pip install command_runner
Usage:
from command_runner.elevate import elevate
def main():
"""My main function that should be elevated"""
print("Who's the administrator, now ?")
if __name__ == '__main__':
elevate(main)
[/EDIT]
import sys,ctypes,platform
def is_admin():
try:
return ctypes.windll.shell32.IsUserAnAdmin()
except:
raise False
if __name__ == '__main__':
if platform.system() == "Windows":
if is_admin():
main(sys.argv[1:])
else:
# Re-run the program with admin rights, don't use __file__ since py2exe won't know about it
# Use sys.argv[0] as script path and sys.argv[1:] as arguments, join them as lpstr, quoting each parameter or spaces will divide parameters
lpParameters = ""
# Litteraly quote all parameters which get unquoted when passed to python
for i, item in enumerate(sys.argv[0:]):
lpParameters += '"' + item + '" '
try:
ctypes.windll.shell32.ShellExecuteW(None, "runas", sys.executable, lpParameters , None, 1)
except:
sys.exit(1)
else:
main(sys.argv[1:])
For one-liners, put the code to where you need UAC.
Request UAC, if failed, keep running:
import ctypes, sys
ctypes.windll.shell32.IsUserAnAdmin() or ctypes.windll.shell32.ShellExecuteW(
None, "runas", sys.executable, " ".join(sys.argv), None, 1) > 32 and exit()
Request UAC, if failed, exit:
import ctypes, sys
ctypes.windll.shell32.IsUserAnAdmin() or (ctypes.windll.shell32.ShellExecuteW(
None, "runas", sys.executable, " ".join(sys.argv), None, 1) > 32, exit())
Function style:
# Created by BaiJiFeiLong#gmail.com at 2022/6/24
import ctypes
import sys
def request_uac_or_skip():
ctypes.windll.shell32.IsUserAnAdmin() or ctypes.windll.shell32.ShellExecuteW(
None, "runas", sys.executable, " ".join(sys.argv), None, 1) > 32 and sys.exit()
def request_uac_or_exit():
ctypes.windll.shell32.IsUserAnAdmin() or (ctypes.windll.shell32.ShellExecuteW(
None, "runas", sys.executable, " ".join(sys.argv), None, 1) > 32, sys.exit())
If your script always requires an Administrator's privileges then:
runas /user:Administrator "python your_script.py"

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