I chrooted directory using following commands:
os.chroot("/mydir")
How to return to directory to previous - before chrooting?
Maybe it is possible to unchroot directory?
SOLUTION:
Thanks to Phihag. I found a solution. Simple example:
import os
os.mkdir('/tmp/new_dir')
dir1 = os.open('.', os.O_RDONLY)
dir2 = os.open('/tmp/new_dir', os.O_RDONLY)
os.getcwd() # we are in 'tmp'
os.chroot('/tmp/new_dir') # chrooting 'new_dir' directory
os.fchdir(dir2)
os.getcwd() # we are in chrooted directory, but path is '/'. It's OK.
os.fchdir(dir1)
os.getcwd() # we came back to not chrooted 'tmp' directory
os.close(dir1)
os.close(dir2)
More info
If you haven't changed your current working directory, you can simply call
os.chroot('../..') # Add '../' as needed
Of course, this requires the CAP_SYS_CHROOT capability (usually only given to root).
If you have changed your working directory, you can still escape, but it's harder:
os.mkdir('tmp')
os.chroot('tmp')
os.chdir('../../') # Add '../' as needed
os.chroot('.')
If chroot changes the current working directory, you can get around that by opening the directory, and using fchdir to go back.
Of course, if you intend to go out of a chroot in the course of a normal program (i.e. not a demonstration or security exploit), you should rethink your program. First of all, do you really need to escape the chroot? Why can't you just copy the required info into it beforehand?
Also, consider using a second process that stays outside of the chroot and answers to the requests of the chrooted one.
Related
I am fairly new to programming and currently working on a python script. It is supposed to gather all the files and directories that are given as paths inside the program and copy them to a new location that the user can choose as an input.
import shutil
import os
from pathlib import Path
import argparse
src = [ [insert name of destination directory, insert path of file/directory that
should be copied ]
]
x = input("Please choose a destination path\n>>>")
if not os.path.exists(x):
os.makedirs(x)
print("Directory was created")
else:
print("Existing directory was chosen")
dest = Path(x.strip())
for pfad in src:
if os.path.isdir(pfad[1]):
shutil.copytree(pfad[1], dest / pfad[0])
elif os.path.isfile(pfad[1]):
pfad1 = Path(dest / pfad[0])
if not os.path.exists(pfad1):
os.makedirs(pfad1)
shutil.copy(pfad[1], dest / pfad[0])
else:
print("An error occured")
print(pfad)
print("All files and directories have been copied!")
input()
The script itself is working just fine. The problem is that I want write a test that automatically test the code each time I push it to my GitLab repository. I have been browsing through the web for quite some time now but wasnt able to find a good explanation on how to approach creating a test for a script like this.
I would be extremely thankful for any kind of feedback or hints to helpful resources.
First, you should write a test that you can run in command line.
I suggest you use the argparse module to pass source and destination directories, so that you can run thescript.py source_dir dest_dir without human interaction.
Then, as you have a test you can run, you need to add a .gitlab-ci.yml to the root of the project so that you can use the gitlab CI.
If you never used the gitlab CI, you need to start here: https://docs.gitlab.com/ee/ci/quick_start/
After that, you'll be able to add a job to your .gitlab-ci.yml, so that a runner with python installed will run the test. If you don't understad the bold terms of the previous sentence, you need to understant Gitlab CI first.
Working with scientific data, specifically climate data, I am constantly hard-coding paths to data directories in my Python code. Even if I were to write the most extensible code in the world, the hard-coded file paths prevent it from ever being truly portable. I also feel like having information about the file system of your machine coded in your programs could be security issue.
What solutions are out there for handling the configuration of paths in Python to avoid having to code them out explicitly?
One of the solution rely on using configuration files.
You can store all your path in a json file like so :
{
"base_path" : "/home/bob/base_folder",
"low_temp_area_path" : "/home/bob/base/folder/low_temp"
}
and then in your python code, you could just do :
import json
with open("conf.json") as json_conf :
CONF = json.load(json_conf)
and then you can use your path (or any configuration variable you like) like so :
print "The base path is {}".format(CONF["base_path"])
First off its always good practise to add a main function to go with each class to test that class or functions in the file. Along with this you determine the current working directory. This becomes incredibly important when running python from a cron job or from a directory that is not the current working directory. No JSON files or environment variables are then needed and you will obtain interoperation across Mac, RHEL and Debian distributions.
This is how you do it, and it will work on windows also if you use '\' instead of '/' (if that is even necessary, in your case).
if "__main__" == __name__:
workingDirectory = os.path.realpath(sys.argv[0])
As you can see when you run your command, the working directory is calculated if you provide a full path or relative path, meaning it will work in a cron job automatically.
After that if you want to work with data that is stored in the current directory use:
fileName = os.path.join( workingDirectory, './sub-folder-of-current-directory/filename.csv' )
fp = open( fileName,'r')
or in the case of the above working directory (parallel to your project directory):
fileName = os.path.join( workingDirectory, '../folder-at-same-level-as-my-project/filename.csv' )
fp = open( fileName,'r')
I believe there are many ways around this, but here is what I would do:
Create a JSON config file with all the paths I need defined.
For even more portability, I'd have a default path where I look for this config file but also have a command line input to change it.
In my opinion passing arguments from command line would be best solution. You should take a look at argparse . This allows you to create nice way to handle arguments from the command line. for example:
myDataScript.py /home/userName/datasource1/
I have a string of Java source code in Python that I want to compile, execute, and collect the output (stdout and stderr). Unfortunately, as far as I can tell, javac and java require real files, so I have to create a temporary directory.
What is the best way to do this? The tempfile module seems to be oriented towards creating files and directories that are only visible to the Python process. But in this case, I need Java to be able to see them too. However, I also want the other stuff to be handled intelligently if possible (such as deleting the folder when done or using the appropriate system temp folder)
tempfile.NamedTemporaryFile and tempfile.TemporaryDirectory work perfectly fine for your purposes. The resulting objects have a .name attribute that provides a file system visible name that java/javac can handle just fine, just make sure to:
Set the suffix appropriately if the compiler insists on files being named with a .java extension
Always call .flush() on the file handle before handing the .name of a NamedTemporaryFile to an external process or it may (usually will) see an incomplete file
If you don't want Python cleaning up the files when you close the objects, either pass delete=False to NamedTemporaryFile's constructor, or use the mkstemp and mkdtemp functions (which create the objects, but don't clean them up for you).
So for example, you might do:
# Create temporary directory for source and class files
with tempfile.TemporaryDirectory() as d:
# Write source code
srcpath = os.path.join(d.name, "myclass.java")
with open(srcpath, "w") as srcfile:
srcfile.write('source code goes here')
# Compile source code
subprocess.check_call(['javac', srcpath])
# Run source code
# Been a while since I've java-ed; you don't include .java or .class
# when running, right?
invokename = os.path.splitext(srcpath)[0]
subprocess.check_call(['java', invokename])
... with block for TemporaryDirectory done, temp directory cleaned up ...
tempfile.mkstemp creates a file that is normally visible in the filesystem and returns you the path as well. You should be able to use this to create your input and output files - assuming javac will atomically overwrite the output file if it exists there should be no race condition if other processes on your system don't misbehave.
I am using django views, I create a temp_dir using tempfile.gettempdir().
I write a gzipped text file in there, and then scp the file elsewhere. When these tasks are complete I try to delete the temp_dir.
if os.path.exists( temp_dir ):
shutil.rmtree( temp_dir )
However, occasionally I get this error back:
Operation not permitted: '/tmp/.ICE-unix'
Any ideas what this error means and how to best handle this situation?
tempfile.gettempdir() does not create a temp directory - it returns your system's standand tmp directory. DO NOT DELETE IT! That will blow everybody's temp files away. You can delete the file you created inside the temp dir, or you can create your own temp dir, but leave this one alone.
The value for temp_dir is taken from the OS environment variables, and apparently some other process is also using it to create files. The other file might be in use/locked and that will prevent you from deleting it.
Q: What is /tmp/.ICE-unix ?
A: Its a directory where X-windows session information is saved.
I am no expert but try running the python program or what your using to do this as an administrator then it will most likely allow this process to be done...
I see that if we change the HOME (linux) or USERPROFILE (windows) environmental variable and run a python script, it returns the new value as the user home when I try
os.environ['HOME']
os.exp
Is there any way to find the real user home directory without relying on the environmental variable?
edit:
Here is a way to find userhome in windows by reading in the registry,
http://mail.python.org/pipermail/python-win32/2008-January/006677.html
edit:
One way to find windows home using pywin32,
from win32com.shell import shell,shellcon
home = shell.SHGetFolderPath(0, shellcon.CSIDL_PROFILE, None, 0)
I think os.path.expanduser(path) could be helpful.
On Unix and Windows, return the argument with an initial component of ~ or ~user replaced by that user‘s home directory.
On Unix, an initial ~ is replaced by the environment variable HOME if it is set; otherwise the current user’s home directory is looked up in the password directory through the built-in module pwd. An initial ~user is looked up directly in the password directory.
On Windows, HOME and USERPROFILE will be used if set, otherwise a combination of HOMEPATH and HOMEDRIVE will be used. An initial ~user is handled by stripping the last directory component from the created user path derived above.
If the expansion fails or if the path does not begin with a tilde, the path is returned unchanged.
So you could just do:
os.path.expanduser('~user')
from pathlib import Path
str(Path.home())
works in Python 3.5 and above. Path.home() returns a Path object providing an API I find very useful.
I think os.path.expanduser(path) is the best answer to your question, but there's an alternative that may be worth mentioning in the Unix world: the pwd package. e.g.
import os, pwd
pwd.getpwuid(os.getuid()).pw_dir
For windows;
import os
homepath = os.path.expanduser(os.getenv('USERPROFILE'))
will give you a handle to current user's home directory and
filepath = os.path.expanduser(os.getenv('USERPROFILE'))+'\\Documents\\myfile.txt'
will give you a handle to below file;
C:\Users\urUserName\Documents\myfile.txt
home_folder = os.getenv('HOME')
This should work on Windows and Mac OS too, works well on Linux.
Really, a change in environment variable indicates that the home must be changed. So every program/script should have the new home in context; also the consequences are up to the person who changed it.
I would still stick with
home = os.getenv('USERPROFILE') or os.getenv('HOME')
what exactly is required?
I realize that this is an old question that has been answered but I thought I would add my two cents. The accepted answer was not working for me. I needed to find the user directory and I wanted it to work with and without sudo. In Linux, my user directory is "/home/someuser" but my root directory is "/root/". However, on my Mac, the user directory is "/Users/someuser". Here is what I ended up doing:
_USERNAME = os.getenv("SUDO_USER") or os.getenv("USER")
_HOME = os.path.expanduser('~'+_USERNAME)
This worked both with and without sudo on Mac and Linux.
get (translated) user folder names on Linux:
from gi.repository import GLib
docs = GLib.get_user_special_dir(GLib.USER_DIRECTORY_DOCUMENTS)
desktop = GLib.get_user_special_dir(GLib.USER_DIRECTORY_DESKTOP)
pics = GLib.get_user_special_dir(GLib.USER_DIRECTORY_PICTURES)
videos = GLib.get_user_special_dir(GLib.USER_DIRECTORY_VIDEOS)
music = GLib.get_user_special_dir(GLib.USER_DIRECTORY_MUSIC)
downloads = GLib.get_user_special_dir(GLib.USER_DIRECTORY_DOWNLOAD)
public = GLib.get_user_special_dir(GLib.USER_DIRECTORY_PUBLIC_SHARE)
templates = GLib.get_user_special_dir(GLib.USER_DIRECTORY_TEMPLATES)
print(docs)
print(desktop)
print(pics)
print(videos)
print(music)
print(downloads)
print(public)
print(templates)
On Linux and other UNIXoids you can always take a peek in /etc/passwd. The home directory is the sixth colon-separated field in there. No idea on how to do better than the environment variable on Windows though. There'll be a system call for it, but if it's available from Python, ...