I am struggling to convert a url to a nested tuple.
# Convert this string
str = 'http://somesite.com/?foo=bar&key=val'
# to a tuple like this:
[(u'foo', u'bar'), (u'key', u'val')]
I assume I need to be doing something like:
url = 'http://somesite.com/?foo=bar&key=val'
url = url.split('?')
get = ()
for param in url[1].split('&'):
get = get + param.split('=')
What am I doing wrong? Thanks!
I believe you are looking for the urlparse module.
This module defines a standard interface to break Uniform Resource Locator (URL) strings up in components (addressing scheme, network location, path etc.), to combine the components back into a URL string, and to convert a “relative URL” to an absolute URL given a “base URL.”
Here is an example:
from urlparse import urlparse, parse_qsl
url = 'http://somesite.com/?foo=bar&key=val'
print parse_qsl(urlparse(url)[4])
Output:
[('foo', 'bar'), ('key', 'val')]
In this example I first use the urlparse function to parse the entire URL then I use the parse_qsl function to break the querystring (the fifth element returned from urlparse) into a list of tuples.
Andrew's answer was really informative and helpful. A less adept way to grab those params would be with a regular expression--something like this:
import re
re_param = re.compile(r'(?P<key>w\+)=(?P<value>w\+)')
url = 'http://somesite.com/?foo=bar&key=val''
params_list = re_param.findall(url)
Also, in your code it looks like you're trying to concatenate a list and tuple--
for param in url[1].split('&'):
get = get + param.split('=')
You created get as a tuple, but str.split returns a list. Maybe this would fix your code:
for param in url[1].split('&'):
get = get + tuple(param.split('='))
