Why does this function fail to read XML from "https://www.seattletimes.com/feed/"?
I can visit the URL from my browser just fine. It also reads XML from other websites without a problem ("https://news.ycombinator.com/rss").
import urllib
def get_url(u):
header = {'User-Agent': 'Mozilla/5.0'}
request = urllib.request.Request(url=url, headers=header)
response = urllib.request.urlopen(request)
return response.read().decode('utf-8')
url = 'https://www.seattletimes.com/feed/'
feed = get_url(url)
print(feed)
The program times out every time.
Ideas?:
- Maybe
headerneed more info (Accept, etc.)?
EDIT1:
I replaced with the request header from the script with my browser header. Still no-go.
header = {
'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9',
'Accept-Encoding': 'gzip, deflate',
'Accept-Language': 'en-US,en;q=0.9',
'Connection': 'keep-alive',
'Accept-Language': 'en-US,en;q=0.9',
'Connection': 'keep-alive',
'Upgrade-Insecure-Requests': '1',
'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/96.0.4664.110 Safari/537.36' }
>Solution :
I am not quite sure why but the header/user-agent was confusing the website. If you remove it your code works just fine. I’ve tried different header arguments without issues, the user-agent seems to be what causes that behaviour.
import urllib.request
def get_url(u):
request = urllib.request.Request(url=url)
response = urllib.request.urlopen(request)
return response.read().decode('utf-8')
url = 'https://www.seattletimes.com/feed/'
feed = get_url(url)
print(feed)