urlparse

This question already has an answer here: no module named urllib.parse (How should I install it?) 11 answers I would like to use urlparse . But python3.4.1 is not finding the module. I do import urlparse , but it gives me this error importError: no 'module' named ulrparse The urlparse in Python 2.7.11 was renamed to urllib.parse in Python 3. So, if you have a code such from urlparse import urljoin , I suggest you change it to from urllib.parse import urljoin As noted in urlparse 's documentation : Note The urlparse module is renamed to urllib.parse in Python 3. The 2to3 tool will automatically

I recently started working on a project using just vim as my text editor with a virtualenv setup. I installed a few API's on this virtualenv from GitHub. Eventually, the project got a little bigger than vim could handle so I had to move the project to an IDE. I chose Aptana Studio 3. When I started up Aptana, I pointed the project directory to the virtualenv folder that I had created to house my project. I then pointed the interpreter at the Python executable in App/bin (created from virtualenv)/python2.7. When I started reworking the code to make sure I had everything mapped correctly, I was

I'd like to add the 'http' scheme name in front of a given url string if it's missing. Otherwise, leave the url alone so I thought urlparse was the right way to do this. But whenever there's no scheme and I use get url, I get /// instead of '//' between the scheme and domain. >>> t = urlparse.urlparse('www.example.com', 'http') >>> t.geturl() 'http:///www.example.com' # three /// How do I convert this url so it actually looks like: 'http://www.example.com' # two // Short answer (but it's a bit tautological): >>> urlparse.urlparse("http://www.example.com").geturl() 'http://www.example.com' In

I need to get the query string from this URL https://stackoverflow.com/questions/ask?next=1&value=3 and I don't want to use request.META . I have figured out that there are two more ways to get the query string: Using urlparse urlparse.urlparse(url).query Using url encode Use urlencode and pass the request.GET params dictionary into it to get the string representation. So which way is better? My colleagues prefer urlencode but have not provided a satisfying explanation. They claim that urlparse calls urlencode internally which is something I'm not sure about since urlencode lives in the urllib

My application creates custom URIs (or URLs?) to identify objects and resolve them. The problem is that Python's urlparse module refuses to parse unknown URL schemes like it parses http. If I do not adjust urlparse's uses_* lists I get this: >>> urlparse.urlparse("qqqq://base/id#hint") ('qqqq', '', '//base/id#hint', '', '', '') >>> urlparse.urlparse("http://base/id#hint") ('http', 'base', '/id', '', '', 'hint') Here is what I do, and I wonder if there is a better way to do it: import urlparse SCHEME = "qqqq" # One would hope that there was a better way to do this urlparse.uses_netloc.append

I've taken a look to urlparse.urlparse method documentation and I'm a little bit confused about what is the parameters part (not to be confused with the more familiar query part, that is what goes after the question mark and before the fragment part). Wikipedia entry on URL's structure doesn't say anything about that, so could please anybody elaborate a little bit on this and possibly give some examples? fascinating, this is the first time I've encounter them, found this http://doriantaylor.com/policy/http-url-path-parameter-syntax I also found this http://tools.ietf.org/html/rfc3986#section-3

I'm trying to write a script to test for the existence of a web page, would be nice if it would check without downloading the whole page. This is my jumping off point, I've seen multiple examples use httplib in the same way, however, every site I check simply returns false. import httplib from httplib import HTTP from urlparse import urlparse def checkUrl(url): p = urlparse(url) h = HTTP(p[1]) h.putrequest('HEAD', p[2]) h.endheaders() return h.getreply()[0] == httplib.OK if __name__=="__main__": print checkUrl("http://www.stackoverflow.com") # True print checkUrl("http://stackoverflow.com

Im looking for a method (or function) to strip out the domain.ext part of any URL thats fed into the function. The domain extension can be anything (.com, .co.uk, .nl, .whatever), and the URL thats fed into it can be anything from http://www.domain.com to www.domain.com/path/script.php?=whatever Whats the best way to go about doing this? Robert Elwell parse_url turns a URL into an associative array: php > $foo = "http://www.example.com/foo/bar?hat=bowler&accessory=cane"; php > $blah = parse_url($foo); php > print_r($blah); Array ( [scheme] => http [host] => www.example.com [path] => /foo/bar