Audit required: External control of file name or path PTC-W6004
External variable 'package_json_path' used in file path
10 """
11
12 # Load the package.json file
13 with open(package_json_path, "r") as file:14 package_json = json.load(file)
15
16 # Get the list of dependenciesExternal variable 'file_path' used in file path
5 config = configparser.ConfigParser()
6
7 # Read the configuration file
8 with open(file_path, "r") as file: 9 config.read_file(file)
10
11 # Extract project settings and parametersExternal variable 'guidelines_path' used in file path
10 """
11
12 # Read the current content of the guidelines document
13 with open(guidelines_path, "r") as file:14 current_guidelines_content = file.read()
15
16 # Replace the old content with the new content
Description
Python's open() function can take in a relative or absolute path and read its file contents.
If a user is provided direct access to the path that is opened, it can have serious security risks.
Bad practice
def read_file(path):
with open(os.path.join('some/path', path)) as f:
f.read()
# Someone can exploit `read_file` and see your secrets this way:
read_file('../../../secrets.txt')
Recommended
Either use a static path:
def read_file(path):
with open('some/path/to/file.txt') as f:
f.read()
Or, do some kind of validation to make sure you're not allowing arbitrary file access:
def read_file(filename):
if filename not in ('x.txt', 'y.txt'):
return 'Invalid filename'
with open(os.path.join('some/path', path)) as f:
f.read()
References
- OWASP Top 10 2021 Category A01 - Broken Access Control
- OWASP Top 10 2021 Category A04 - Insecure Design
- CWE-73 External Control of File Name or Path