get http status code python selenium

Question

How can I get the http status code in python if I open a link? I did it this way, but it always shows 200 even if the page throws a 404 error...

from urllib.request import *

self.driver.get(link)
code = urlopen(link).getcode()
if not code == 200: 
  print('http status: ' + str(code))

posible dublicate https://stackoverflow.com/questions/5799228/how-to-get-status-code-by-using-selenium-py-python-code — Druta Ruslan, May 18 '18 at 09:37

score 1 · Accepted Answer · answered May 18 '18 at 09:40

1

That's the problem, Selenium will always find a page. You have 2 ways to check if loading didn't work :

1 : Check content

self.assertFalse("<insert here a string which is only in the error page>" in self.driver.page_source)

2 : Use Django test Client

response = self.client.get(link)
self.assertEqual(response.status_code,200)

answered May 18 '18 at 09:40

AntoineLB

482
3
19

ok thanks, actually I can't get django work...it has some issues with the settings – marshmallow May 18 '18 at 11:34
If you need some help about it i'm there. – AntoineLB May 24 '18 at 15:42

score 0 · Answer 2 · answered May 18 '18 at 12:06

0

Edited your example. Try to use requests library.

from requests import get

self.driver.get(link)
request = get(link)
if not request.status_code == 200:
  print('http status: ' + str(request.status_code))

Hope, this will help you.

answered May 18 '18 at 12:06

Oleksandr Makarenko

779
1
6
18

get http status code python selenium

2 Answers2