Pythonic URL Parsing

Question

There are a number of questions about how to parse a URL in Python, this question is about the best or most Pythonic way to do it.

In my parsing I need 4 parts: the network location, the first part of the URL, the path and the filename and querystring parts.

http://www.somesite.com/base/first/second/third/fourth/foo.html?abc=123

should parse into:

netloc = 'www.somesite.com'
baseURL = 'base'
path = '/first/second/third/fourth/'
file = 'foo.html?abc=123'

The code below produces the correct result, but is there are better way to do this in Python?

url = "http://www.somesite.com/base/first/second/third/fourth/foo.html?abc=123"

file=  url.rpartition('/')[2]
netloc = urlparse(url)[1]
pathParts = path.split('/')
baseURL = pathParts[1]

partCount = len(pathParts) - 1

path = "/"
for i in range(2, partCount):
    path += pathParts[i] + "/"


print 'baseURL= ' + baseURL
print 'path= ' + path
print 'file= ' + file
print 'netloc= ' + netloc

Exact Duplicate: stackoverflow.com/questions/258746/slicing-url-with-python — S.Lott
– S.Lott, Commented May 19, 2009 at 10:47
Not quite the same as 258746, this question had a slightly different goal and the main focus of asking was about the best (Pythonic) way to accomplish the task. — Shawn Swaner
– Shawn Swaner, Commented May 20, 2009 at 8:17

Paolo Bergantino · Accepted Answer · 2009-05-19 04:44:12Z

6

Since your requirements on what parts you want are different from what urlparse gives you, that's as good as it's going to get. You could, however, replace this:

partCount = len(pathParts) - 1

path = "/"
for i in range(2, partCount):
    path += pathParts[i] + "/"

With this:

path = '/'.join(pathParts[2:-1])

answered May 19, 2009 at 4:44

Paolo Bergantino

490k83 gold badges523 silver badges437 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Laurence Gonsalves · Accepted Answer · 2009-05-19 21:51:37Z

2

I'd be inclined to start out with urlparse. Also, you can use rsplit, and the maxsplit parameter of split and rsplit to simplify things a bit:

_, netloc, path, _, q, _ = urlparse(url)
_, base, path = path.split('/', 2) # 1st component will always be empty
path, file = path.rsplit('/', 1)
if q: file += '?' + q

edited May 19, 2009 at 21:51

answered May 19, 2009 at 6:55

Laurence Gonsalves

144k38 gold badges264 silver badges315 bronze badges

Collectives™ on Stack Overflow

Pythonic URL Parsing

2 Answers 2

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related