I'm doing this:
urlparse.urljoin('http://example.com/mypage', '?name=joe')
And I get this:
'http://example.com/?name=joe'
While I want to get this:
'http://example.com/mypage?name=joe'
What am I doing wrong?
I'm doing this:
urlparse.urljoin('http://example.com/mypage', '?name=joe')
And I get this:
'http://example.com/?name=joe'
While I want to get this:
'http://example.com/mypage?name=joe'
What am I doing wrong?
I solved it by bundling Python 2.6's urlparse
module with my project. I also had to bundle namedtuple
which was defined in collections
, since urlparse
uses it.
You could use urlparse.urlunparse :
import urlparse
parsed = list(urlparse.urlparse('http://example.com/mypage'))
parsed[4] = 'name=joe'
urlparse.urlunparse(parsed)
You're experiencing a known bug which affects Python 2.4-2.6.
If you can't change or patch your version of Python, @jd's solution will work around the issue.
However, if you need a more generic solution that works as a standard urljoin
would, you can use a wrapper method which implements the workaround for that specific use case, and default to the standard urljoin()
otherwise.
For example:
import urlparse
def myurljoin(base, url, allow_fragments=True):
if url[0] != "?":
return urlparse.urljoin(base, url, allow_fragments)
if not allow_fragments:
url = url.split("#", 1)[0]
parsed = list(urlparse.urlparse(base))
parsed[4] = url[1:] # assign params field
return urlparse.urlunparse(parsed)
I solved it by bundling Python 2.6's urlparse
module with my project. I also had to bundle namedtuple
which was defined in collections
, since urlparse
uses it.
Are you sure? On Python 2.7:
>>> import urlparse
>>> urlparse.urljoin('http://example.com/mypage', '?name=joe')
'http://example.com/mypage?name=joe'
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