Python: How to go to a link and open it if it exists

Question

I found this code:

import webbrowser

webbrowser.open('http://eample.com')

How do I modify it so that it will only open the link if the return code is 200?

Answer 1

Use the requests package to send a HEAD request to the desired location and then create an if

if requests.head(url).status_code == 200: webbrowser.open(url) # open url if status is 200