Reading a "special" URL - python-3.x

Question

The task is simple- I want to tranfer a html file from an URL to a variable and read the feed below:
How can I read the contents of an URL with Python?
All that works well except with the url = "https://www.goyax.de/"
with
import urllib
#fp = urllib.request.urlopen("https://www.spiegel.de/")
fp = urllib.request.urlopen("https://www.goyax.de/")
print("Result code: " + str(fp.getcode()))
print("Returned data: -----------------")
data = fp.read().decode("utf-8")
print(data)
I get only "403" and "Forbidden". Also with
import requests
url = 'https://www.goyax.de/'
#url = 'https://www.spiegel.de'
r = requests.get(url)
tt = r.text
print(tt)
I don't get an improvement. With other URLs both solutions work well so far.
Until now I was using an Autohotkey script (UrlDownloadToFile) (Windows only) and tried it also with Octave (s = urlread("https://www.goyax.de/")) where I get the right result and no error message. the scripts ae running sicne years on a PC but I want to move this task to a Raspberry Pi. Because of that I was learning Python
The output / error messages:
fp = urllib.request.urlopen("http://www.goyax.de/")
File "C:\ProgramData\Anaconda3\lib\urllib\request.py", line 222, in urlopen
return opener.open(url, data, timeout)
File "C:\ProgramData\Anaconda3\lib\urllib\request.py", line 531, in open
response = meth(req, response)
File "C:\ProgramData\Anaconda3\lib\urllib\request.py", line 640, in http_response
response = self.parent.error(
File "C:\ProgramData\Anaconda3\lib\urllib\request.py", line 563, in error
result = self._call_chain(*args)
File "C:\ProgramData\Anaconda3\lib\urllib\request.py", line 502, in _call_chain
result = func(*args)
File "C:\ProgramData\Anaconda3\lib\urllib\request.py", line 755, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "C:\ProgramData\Anaconda3\lib\urllib\request.py", line 531, in open
response = meth(req, response)
File "C:\ProgramData\Anaconda3\lib\urllib\request.py", line 640, in http_response
response = self.parent.error(
File "C:\ProgramData\Anaconda3\lib\urllib\request.py", line 569, in error
return self._call_chain(*args)
File "C:\ProgramData\Anaconda3\lib\urllib\request.py", line 502, in _call_chain
result = func(*args)
File "C:\ProgramData\Anaconda3\lib\urllib\request.py", line 649, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
HTTPError: Forbidden

Answer 1

Well, I found the answer myself or with some help of a friend: The key is to the soluition is to set the user agent.
Solution 1) (with "Requests")
import requests
r = requests.get(url2, headers={"User-Agent":'Mozilla/5.0'}, timeout=25)
mystr=r.text # mybytes=r.content
print(mystr)
Solution 2) (with "urllib.request" + "CookieJar")
import urllib.request
from http.cookiejar import CookieJar
req = urllib.request.Request(url2, None, {"User-Agent":'Mozilla/5.0'})
# instead of #req = urllib.request.Request(url2)
opener = urllib.request.build_opener(urllib.request.HTTPCookieProcessor(CookieJar()))
response = opener.open(req)
content = response.read()
mystr2 = content.decode("utf8")
print(mystr2)
Usually the user agent 'Mozilla/5.0' is sufficient. Or check
https://manytools.org/http-html-text/user-agent-string/
and
https://www.scrapehero.com/how-to-fake-and-rotate-user-agents-using-python-3/
for a real user agent string.