How do I get a content-type of a file in Python? (with url..)

Question

Suppose I haev a video file:

http://mydomain.com/thevideofile.mp4

How do I get the header and the content-type of this file? With Python. But , I don't want to download the entire file. i want it to return:

video/mp4

Edit: this is what I did. What do you think?

f = urllib2.urlopen(url)
    params['mime'] =  f.headers['content-type']

Answer 1

Like so:

>>> import httplib
>>> conn = httplib.HTTPConnection("mydomain.com")
>>> conn.request("HEAD", "/thevideofile.mp4")
>>> res = conn.getresponse()
>>> print res.getheaders()

That will only download and print the headers because it is making a HEAD request:

Asks for the response identical to the one that would correspond to a GET request, but without the response body. This is useful for retrieving meta-information written in response headers, without having to transport the entire content.

(via Wikipedia)

Answer 2

This is a higher level answer than Brian's. Using the urllib machinery has the usual advantages such as handling redirects automatically and so on.

import urllib2

class HeadRequest(urllib2.Request):
    def get_method(self):
        return "HEAD"

url = "http://mydomain.com/thevideofile.mp4"
head = urllib2.urlopen(HeadRequest(url))
head.read()          # This will return empty string and closes the connection
print head.headers.maintype
print head.headers.subtype
print head.headers.type