How can I prepend the 'http://' protocol to a url when necessary?

Question

I need to parse an URL. I'm currently using urlparse.urlparse() and urlparse.urlsplit().

The problem is that i can't get the "netloc" (host) from the URL when it's not present the scheme. I mean, if i have the following URL:

www.amazon.com/Programming-Python-Mark-Lutz/dp/0596158106/ref=sr_1_1?ie=UTF8&qid=1308060974&sr=8-1

I can't get the netloc: www.amazon.com

According to python docs:

Following the syntax specifications in RFC 1808, urlparse recognizes a netloc only if it is properly introduced by ‘//’. Otherwise the input is presumed to be a relative URL and thus to start with a path component.

So, it's this way on purpose. But, i still don't know how to get the netloc from that URL.

I think i could check if the scheme is present, and if it's not, then add it, and then parse it. But this solution doesn't seems really good.

Do you have a better idea?

EDIT: Thanks for all the answers. But, i cannot do the "startswith" thing that's proposed by Corey and others. Becouse, if i get an URL with other protocol/scheme i would mess it up. See:

If i get this URL:

ftp://something.com

With the code proposed i would add "http://" to the start and would mess it up.

The solution i found

if not urlparse.urlparse(url).scheme:
   url = "http://"+url
return urlparse.urlparse(url)

Something to note:

I do some validation first, and if no scheme is given i consider it to be http://

Answer 1

looks like you need to specify the protocol to get netloc.

adding it if it's not present might look like this:

import urlparse

url = 'www.amazon.com/Programming-Python-Mark-Lutz'
if '//' not in url:
    url = '%s%s' % ('http://', url)
p = urlparse.urlparse(url)
print p.netloc