Using Beautiful Soup to get the full URL in source code
python
Solution
<a href="/folder/big/a.jpg">
That’s an absolute address for the current host. So if the HTML file is at `http://example.com/foo/bar.html`, then applying the url `/folder/big/a.jpg` will result in this:
http://example.com/folder/big/a.jpg
I.e. take the host name and apply the new path to it.
Python has the builtin `urljoin` function to perform this operation for you:
>>> from urllib.parse import urljoin
>>> base = 'http://example.com/foo/bar.html'
>>> href = '/folder/big/a.jpg'
>>> urljoin(base, href)
'http://example.com/folder/big/a.jpg'
For Python 2, the function is within the `urlparse` module.
Problem
So I was looking at some source code and I came across this bit of code ``` <img src="/gallery/2012-winners-finalists/HM_Watching%20birds2_Shane%20Conklin_MA_2012.jpg" ``` now in the source code the link is blue and when you click it, it takes you to the full URL where that picture is located, I know how to get what is shown in the source code in Python using Beautiful Soup I was wondering though how to get the full URL you get once clicking the link in the source code? EDIT: if I was given `<a href = "/folder/big/a.jpg"` how do you figure out the starting part of that url through python or beautiful soup?