(Title might change not too sure what to call it)
So I'm trying to open a URL that directs to a random page (This URL: http://anidb.net/perl-bin/animedb.pl?show=anime&do.random=1) and I want to return where that URL goes
randomURL = urllib.urlopen("http://anidb.net/perl-bin/animedb.pl?show=anime&do.random=1")
print(randomURL)
That's what I (stupidly) thought would work. I imported urllib
In Python 3 >, urllib.urlopen
was replaced by urllib.request.urlopen
. Change the request line to this:
urllib.request.urlopen('http://anidb.net/perl-bin/animedb.pl?show=anime&do.random=1')
For more, you can see the docs
But if you want to have the url, which is a bit more difficult, you can take a look at urllib.request.HTTPRedirectHandler