Printing first line then assign to a variable

for a in soup.find_all('a', href=True):
    if "/xxxxxx/xxxxxxxxxxxxx-" in a['href']:
        x = ("https://www.unkown.com" + a['href'])
        print(x[0])
    else:
        pass
        
my output:
https://www.unkown.com/xxxxx/xxxxxxxxxx
https://www.unkown.com/xxxxx/1xxxxxxxxx
https://www.unkown.com/xxxxx/2xxxxxxxxx
https://www.unkown.com/xxxxx/3xxxxxxxxx

I want to print the first line by doing print(x[0]) I just print "h" the output is bunch of url's

If your output x is :

https://www.unkown.com/xxxxx/xxxxxxxxxx
https://www.unkown.com/xxxxx/1xxxxxxxxx
https://www.unkown.com/xxxxx/2xxxxxxxxx
https://www.unkown.com/xxxxx/3xxxxxxxxx

You can access the first line by calling x.split('\n')[0].

In Python, string indexing (x[0]) returns the character at that index, not an element from a list. In your case, x is a string, so x[0] will return 'h', the first character of your URL.

To print only the first matching URL, you can use a boolean flag. Here's an example:

first_match_found = False

for a in soup.find_all('a', href=True):
    if "/xxxxxx/xxxxxxxxxxxxx-" in a['href'] and not first_match_found:
        print("https://www.unkown.com" + a['href'])
        first_match_found = True

This script will stop printing URLs after the first match.