title: Detect DOM Based XSS website using python
body: <p>I have a sample url: <code>http://www.foo.bar/index.php?ids=111</code>
  <p>I've found that it has a vulnerability by <code>http://www.foo.bar/index.php?ids="&gt;&lt;SCrIpT&gt;alert('XSS')&lt;/ScRiPt&gt;</code></p>
  
  <p>A basic different from normal url and xss url is a alert box has been showed in the browser. But when using python, how can i we see that or detect which url is vulnerability?</p>
  
  <p>I've tried python request library: </p>
  
  <pre><code>import requests
  
  xss_url = 'http://www.foo.bar/index.php?ids="&gt;&lt;SCrIpT&gt;alert('XSS')&lt;/ScRiPt&gt;'
  r = requests.get(xss_url);
  print r.text
  
  normal_url = 'http://www.foo.bar/index.php?ids=111'
  t = requests.get(normal_url);
  print t.text
  </code></pre>
  
  <p>Result of r.text and t.text is the same.</p>