How can I prepend the 'http://' protocol t

Question

This question already has an answer here:

• How can I prepend http to a url if it doesn't begin with http? 2 answers

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

Have you considered just checking for the presence of "http://" at the start of the url, and add it if it's not there? Another solution, assuming that first part really is the netloc and not part of a relative url, is to just grab everything up to the first "/" and use that as the netloc.

Answer 2

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

Answer 3

From the 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 you can just do:

In [1]: from urlparse import urlparse

In [2]: def get_netloc(u):
   ...:     if not u.startswith('http'):
   ...:         u = '//' + u
   ...:     return urlparse(u).netloc
   ...: 

In [3]: get_netloc('www.amazon.com/Programming-Python-Mark-Lutz/dp/0596158106/ref=sr_1_1?ie=UTF8&qid=1308060974&sr=8-1')
Out[3]: 'www.amazon.com'

In [4]: get_netloc('http://www.amazon.com/Programming-Python-Mark-Lutz/dp/0596158106/ref=sr_1_1?ie=UTF8&qid=1308060974&sr=8-1')
Out[4]: 'www.amazon.com'

In [5]: get_netloc('https://www.amazon.com/Programming-Python-Mark-Lutz/dp/0596158106/ref=sr_1_1?ie=UTF8&qid=1308060974&sr=8-1')
Out[5]: 'www.amazon.com'

Answer 4

This one liner would do it.

netloc = urlparse('//' + ''.join(urlparse(url)[1:])).netloc

Answer 5

The documentation has this exact example, just below the text you pasted. Adding '//' if it's not there will get what you want. If you don't know whether it'll have the protocol and '//' you can use a regex (or even just see if it already contains '//') to determine whether or not you need to add it.

Your other option would be to use split('/') and take the first element of the list it returns, which will ONLY work when the url has no protocol or '//'.

EDIT (adding for future readers): a regex for detecting the protocol would be something like re.match('(?:http|ftp|https)://', url)

Answer 6

If the protocol is always http you can use only one line:

return "http://" + url.split("://")[-1]

A better option is to use the protocol if it passed:

return url if "://" in url else "http://" + url