The Python Flask-Security-Too package is used for adding security features to your Flask application. It is an is a independently maintained version of Flask-Security based on the 3.0.0 version of Flask-Security. In Flask-Security-Too from version 3.3.0 and before version 3.4.5, the /login and /change endpoints can return the authenticated users authentication token in response to a GET request. Since GET requests arent protected with a CSRF token, this could lead to a malicious 3rd party site acquiring the authentication token. Version 3.4.5 and version 4.0.0 are patched. As a workaround, if you arent using authentication tokens - you can set the SECURITY_TOKEN_MAX_AGE to 0 (seconds) which should make the token unusable.
Weakness
The web application does not, or can not, sufficiently verify whether a well-formed, valid, consistent request was intentionally provided by the user who submitted the request.
Affected Software
| Name |
Vendor |
Start Version |
End Version |
| Flask-security-too |
Flask-security-too_project |
3.3.0 (including) |
3.4.5 (excluding) |
Potential Mitigations
- Use a vetted library or framework that does not allow this weakness to occur or provides constructs that make this weakness easier to avoid.
- For example, use anti-CSRF packages such as the OWASP CSRFGuard. [REF-330]
- Another example is the ESAPI Session Management control, which includes a component for CSRF. [REF-45]
- Use the “double-submitted cookie” method as described by Felten and Zeller:
- When a user visits a site, the site should generate a pseudorandom value and set it as a cookie on the user’s machine. The site should require every form submission to include this value as a form value and also as a cookie value. When a POST request is sent to the site, the request should only be considered valid if the form value and the cookie value are the same.
- Because of the same-origin policy, an attacker cannot read or modify the value stored in the cookie. To successfully submit a form on behalf of the user, the attacker would have to correctly guess the pseudorandom value. If the pseudorandom value is cryptographically strong, this will be prohibitively difficult.
- This technique requires Javascript, so it may not work for browsers that have Javascript disabled. [REF-331]
References