[Python] urllib not working with api request

im trying to write this code where I request this information from openweathermap.org thourgh an api and try to print the temperature and location at the current time.

Most of the code is kind off a mix of the tips I could find on the internet.

Now I get this error and im stuck. Can anyone help me on the right path again?

Heres my code:

import urllib.request, urllib.parse, urllib.error
import json

while True:
zipcode = input('Enter zipcode: ')
if len(zipcode) < 1: break

url = 'http://api.openweathermap.org/data/2.5/weather?
zip='+zipcode+',nl&appid=db071ece9a338a36e9d7a660ec4f0e37?'

print('Retrieving', url)
uh = urllib.request.urlopen(url)
data = uh.read().decode()
print('Retrieved', len(data), 'characters')

try:
js = json.loads(data)
except:
js = None

temp = js["main"]["temp"]
loc = js["name"]

print("temperatuur:", temp)
print("locatie:", loc)


So the url is this: http://api.openweathermap.org/data/2.5/weather?zip=3032,nl&appid=db071ece9a338a36e9d7a660ec4f0e37

The error im getting is:

Enter zipcode: 3343 Retrieving http://api.openweathermap.org/data/2.5/weather?zip=3343,nl&appid=db071ece9a338a36e9d7a660ec4f0e37? Traceback (most recent call last): File "weatherapi2.py", line 12, in uh = urllib.request.urlopen(url) File "C:\Users\ErfanNariman\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 223, in urlopen return opener.open(url, data, timeout) File "C:\Users\ErfanNariman\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 532, in open response = meth(req, response) File "C:\Users\ErfanNariman\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 642, in http_response 'http', request, response, code, msg, hdrs) File "C:\Users\ErfanNariman\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 570, in error return self._call_chain(*args) File "C:\Users\ErfanNariman\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 504, in _call_chain result = func(*args) File "C:\Users\ErfanNariman\AppData\Local\Programs\Python\Python36\lib\urllib\request.py", line 650, in http_error_default raise HTTPError(req.full_url, code, msg, hdrs, fp) urllib.error.HTTPError: HTTP Error 401: Unauthorized
​

Continue reading...