Python Mechanize, how to get URL parameters

Question

I'm currently working on website scraper. Because I have to log in to access the website, a session ID has to be generated and saved for further usage.

The session ID is at the end of the URL.

https://example.com/something.php?sid=123456789

I tried using the geturl() command but it only returns the URL without any parameters.

What would be the best way to get the url parameters?

ozturkib · Accepted Answer · 2020-11-02 13:59:54Z

0

from urllib.parse import urlparse

parsed = urlparse(url)
print(parsed)

The output:

ParseResult(scheme='https', netloc='example.com', path='/something.php', params='', query='sid=123456789', fragment='')

Then, you can access:

print(parsed.query)

The output:

sid=123456789

Then, you can extract:

sid = parsed.query.split('sid=')[-1]
print(sid)

The output:

123456789

answered Nov 2, 2020 at 13:59

ozturkib

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