Python + BeautifulSoup: How to get ‘href’ attribute of ‘a’ element?

Question

I have the following:

  html =
  \'\'\'

    
      


        

      

Answer 1

You could solve this with just a couple lines of gazpacho:

from gazpacho import Soup

html = """\

    
      
File One
    
    

      Down
    
  
"""

soup = Soup(html)
soup.find("a", {"class": "file-link"}).attrs['href']

Which would output:

'/file-one/additional'