How can i open url in python like php

Question

I have view that perform particular action after that return i just want to call url and that url will open in the same page

def googleredirect(url):
    return (url)

like in PHP :

header('Location: http://www.example.com/');

side note: In PHP, it's a header redirect, instead of "open url" — Raptor, Nov 04 '14 at 06:55
urlopen(); in python function get the response for the function and then print the object. — Vijay, Nov 04 '14 at 07:00

score 2 · Answer 1 · answered Nov 04 '14 at 07:28

If you want to only redirect then use:

# in views.py file
from django.shortcuts import redirect

def some_page(request):
    return redirect('http://www.example.com/')

But if you want to open an URL for screen scraping or some thing else, then use python's own library urllib3 or PyQuery for more better DOM element search

score 0 · Answer 2 · edited May 23 '17 at 11:58

0

see here in django

or in python

redirect("http://www.example.com/");

edited May 23 '17 at 11:58

Community

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answered Nov 04 '14 at 07:04

user3510665

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score 0 · Answer 3 · answered Nov 04 '14 at 07:04

If you mean Django-app (@Raptor removed django tag), your view should return an instance of django's HttpResponse

from django.http import HttpResponseRedirect
from django.core.urlresolvers import reverse

def google_redirect(request):
    ...
    return HttpResponseRedirect(reverse('app_name:url_name'))

How can i open url in python like php

3 Answers3