title: httplib in Python to get the status code...but it is too tricky?
body: <pre><code>&gt;&gt;&gt; import httplib
  &gt;&gt;&gt; conn = httplib.HTTPConnection("www.google.com")
  &gt;&gt;&gt; conn.request("HEAD", "/index.html")
  &gt;&gt;&gt; res = conn.getresponse()
  &gt;&gt;&gt; print res.status, res.reason
  200 OK
  </code></pre>
  
  <p>This code will get the HTTP status code.  However, notice that I split up "google.com" and "/index.html" on 2 lines.</p>
  
  <p>And it's confusing.</p>
  
  <p>What if I want to find the status code of just a general URL???</p>
  
  <pre><code>http://mydomain.com/sunny/boo.avi
  http://anotherdomain.com/podcast.mp3
  http://anotherdomain.com/rss/fee.xml
  </code></pre>
  
  <p>Can't I just stick the URL into it, and make it work?</p>
  
  <p>Edit...I cannot use urllib, because I don't want to downlaod the file</p>