Cross-site request forgery (CSRF) vulnerability in the login interface in MediaWiki 1.15 before 1.15.4 and 1.16 before 1.16 beta 3 allows remote attackers to hijack the authentication of users for requests that (1) create accounts or (2) reset passwords, related to the Special:Userlogin form.
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 |
Mediawiki |
Mediawiki |
1.15.0 (including) |
1.15.0 (including) |
Mediawiki |
Mediawiki |
1.15.0-rc1 (including) |
1.15.0-rc1 (including) |
Mediawiki |
Mediawiki |
1.15.1 (including) |
1.15.1 (including) |
Mediawiki |
Mediawiki |
1.15.2 (including) |
1.15.2 (including) |
Mediawiki |
Mediawiki |
1.15.3 (including) |
1.15.3 (including) |
Mediawiki |
Mediawiki |
1.16.0 (including) |
1.16.0 (including) |
Mediawiki |
Mediawiki |
1.16.0-beta1 (including) |
1.16.0-beta1 (including) |
Mediawiki |
Mediawiki |
1.16.0-beta2 (including) |
1.16.0-beta2 (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