Redash is a package for data visualization and sharing. In Redash version 10.0 and prior, the implementation of Google Login (via OAuth) incorrectly uses the state parameter to pass the next URL to redirect the user to after login. The state parameter should be used for a Cross-Site Request Forgery (CSRF) token, not a static and easily predicted value. This vulnerability does not affect users who do not use Google Login for their instance of Redash. A patch in the master and release/10.x.x branches addresses this by replacing Flask-Oauthlib with Authlib which automatically provides and validates a CSRF token for the state variable. The new implementation stores the next URL on the user session object. As a workaround, one may disable Google Login to mitigate the vulnerability.
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 |
| Redash |
Redash |
* |
10.0.0 (including) |
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