Does my code prevent directory traversal?

问题:

Is the following code snippet from a Python WSGI app safe from directory traversal? It reads a file name passed as parameter and returns the named file.

file_name = request.path_params["file"]
file = open(file_name, "rb")
mime_type = mimetypes.guess_type(file_name)[0]
start_response(status.OK, [('Content-Type', mime_type)])
return file

I mounted the app under http://localhost:8000/file/{file} and sent requests with the URLs http://localhost:8000/file/../alarm.gif and http://localhost:8000/file/%2e%2e%2falarm.gif. But none of my attempts delivered the (existing) file. So is my code already safe from directory traversal?

New approach

It seems like the following code prevents directory traversal:

file_name = request.path_params["file"]
absolute_path = os.path.join(self.base_directory, file_name)
normalized_path = os.path.normpath(absolute_path)

# security check to prevent directory traversal
if not normalized_path.startswith(self.base_directory):
    raise IOError()

file = open(normalized_path, "rb")
mime_type = mimetypes.guess_type(normalized_path)[0]
start_response(status.OK, [('Content-Type', mime_type)])
return file

回答1:

Your code does not prevent directory traversal. You can guard against this with the os.path module.

>>> import os.path
>>> os.curdir
'.'
>>> startdir = os.path.abspath(os.curdir)
>>> startdir
'/home/jterrace'

startdir is now an absolute path where you don't want to allow the path to go outside of. Now let's say we get a filename from the user and they give us the malicious /etc/passwd.

>>> filename = "/etc/passwd"
>>> requested_path = os.path.relpath(filename, startdir)
>>> requested_path
'../../etc/passwd'
>>> requested_path = os.path.abspath(requested_path)
>>> requested_path
'/etc/passwd'

We have now converted their path into an absolute path relative to our starting path. Since this wasn't in the starting path, it doesn't have the prefix of our starting path.

>>> os.path.commonprefix([requested_path, startdir])
'/'

You can check for this in your code. If the commonprefix function returns a path that doesn't start with startdir, then the path is invalid and you should not return the contents.

---

The above can be wrapped to a static method like so:

import os 

def is_directory_traversal(file_name):
    current_directory = os.path.abspath(os.curdir)
    requested_path = os.path.relpath(file_name, start=current_directory)
    requested_path = os.path.abspath(requested_path)
    common_prefix = os.path.commonprefix([requested_path, current_directory])
    return common_prefix != current_directory

回答2:

Use only the base name of the user inputed file:

file_name = request.path_params["file"]
file_name = os.path.basename(file_name)
file = open(os.path.join("/path", file_name), "rb")

os.path.basename strips ../ from the path:

>>> os.path.basename('../../filename')
'filename'

回答3:

There's a much simpler solution here:

relative_path = os.path.relpath(path, start=self.test_directory)
has_dir_traversal = relative_path.startswith(os.pardir)

relpath takes care of normalising path for us. And if the relative path starts with .., then you don't allow it.