Audit required: External control of file name or path PTC-W6004
External variable 'path' used in file path
110 encoding = http_context.query.get('encoding', None)
111 if encoding:
112 content = content.decode('utf-8')
113 with open(path, 'w') as f:114 f.write(content)
115 except OSError as e:
116 raise EndpointError(e)External variable 'template_path' used in file path
109
110 content = {'plain':'', 'html':''}
111
112 with open(template_path, 'r') as h:113 template = Template(h.read())
114 html = template.render(vars)
115
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