Answers for "open link python 3"

4

urllib python

#Used to make requests
import urllib.request

x = urllib.request.urlopen('https://www.google.com/')
print(x.read())
Posted by: Guest on June-26-2020
-1

with urllib.request.urlopen("https://

import urllib.request

req = urllib.request.Request('http://www.voidspace.org.uk')
with urllib.request.urlopen(req) as response:
   the_page = response.read()
Posted by: Guest on April-16-2020

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