How to extract string which contains matched string

Question

My code is below which extract all the elements

for link in soup.find_all('a', href=True):
    print(link['href'])

Output

https://www.example.com/author/1/
https://www.example.com/about/2/
https://www.example.com/author/3/

type of (link['href'])

<cls str>
<cls str>
<cls str>

I need to extract the url which contains 'about'

I tried with print(link['href'] if 'about' in link) which thrown eror

My expected out

https://www.example.com/about/2/

Answer 1

A conditional expression requires an else clause to specify what the expression should return when the condition is false.

But you don't want to print anything in that case. So use an if statement around the print() call.

for link in soup.find_all('a', href=True):
    if 'about' in link['href']:
        print(link['href'])

You could also do the matching in the soup.find_all() call.

for link in soup.find_all('a', href=re.compile(r'about')):
    print(link['href'])

Answer 2

You are searching for 'about' in link whereas it seems the word about is found in link['href']. So try to update the if condition as below.

print(link['href'] if 'about' in link['href'] else '')