not accept a text without www

Question

I am trying to get input from the user, more specifically a link. The 'https://' part is automatic, but I want the user to be asked again if he doesn't put 'www' in the beginning of the link.

This is the code by now:

import requests
from bs4 import BeautifulSoup

linkraw = input("Site-ul: ")

while linkraw[0] != "www":
    print("Pune linkul incepand cu 'www' 2")
    linkraw = input("Site-ul: ")

link = f'https://{linkraw}'

httprequest = requests.get(link)

if httprequest.status_code == 200:
    print("Valid")
if httprequest.status_code != 200:
    print("Invalid")

src = httprequest.content
soup = BeautifulSoup(src,'lxml')

print(soup.find_all('a'))

your assumptions is invalid! `linkraw[0]` is just indexing the first char inserted! what about if the user inserted blank line or just single char or even without site ext _ie_ .com .co or whatever, too much errors! — αԋɱҽԃ αмєяιcαη, Apr 21 '21 at 21:02
Does this answer your question? [How to validate a url in Python? (Malformed or not)](https://stackoverflow.com/questions/7160737/how-to-validate-a-url-in-python-malformed-or-not) — αԋɱҽԃ αмєяιcαη, Apr 21 '21 at 21:03

Prune · Answer 1 · 2021-04-21T21:19:21.087

2

You want to check whether the input begins with "www".

if linkraw.startswith("www"):

Cu placere ... :-)

Your commented case is incorrect:

>>> linkraw = "w.google.com"
>>> linkraw.startswith("www")
False

edited Apr 21 '21 at 21:19

answered Apr 21 '21 at 21:03

Prune

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If I just pun 'w' like in 'w.google.com' it still works and I just want it to work with 'www' – David Mustea Apr 21 '21 at 21:15

score 1 · Answer 2 · answered Apr 21 '21 at 21:11

1

You can check if the first 3 letters of the value are the ones you are looking for.

if linkraw[0:3] == "www"

answered Apr 21 '21 at 21:11

dharmey

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This works, but imho `string.startswith` is clearer – jkr Apr 21 '21 at 21:17

not accept a text without www

2 Answers2