HTTP error 404

[Karin]

hi there

I was trying to get address by using API, but I got HTTP Error. Does it mean URL I code didn't make sense or only I couldn't get data from the URL?

My code is here.

import urllib.request
import urllib.parse

API = "https://api.aoikujira.com/zip/xml.get.php"

#パラメータをURLエンコードする
values = {
    'fmt' : 'xml',     
    'zn' : '1740064'    
}
params = urllib.parse.urlencode(values)

#リクエスト用のURLを作成
url = API + "?" + params
print("url = ", url)

#ダウンロード
data = urllib.request.urlopen(url).read()
text = data.decode("utf-8")
print(text)

And the Error is this

Error:
url =  https://api.aoikujira.com/zip/xml.get.php?fmt=xml&zn=1740064
Traceback (most recent call last):
  File "download-zip.py", line 18, in <module>
    data = urllib.request.urlopen(url).read()
  File "C:\Users\…\Python\Python36-32\lib\urllib\request.py", line 223, in urlopen
    return opener.open(url, data, timeout)
  File "C:\Users\…\Python\Python36-32\lib\urllib\request.py", line 532, in open
    response = meth(req, response)
  File "C:\Users\…\Python\Python36-32\lib\urllib\request.py", line 642, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Users\…\Python\Python36-32\lib\urllib\request.py", line 570, in error
    return self._call_chain(*args)
  File "C:\Users\…\Python\Python36-32\lib\urllib\request.py", line 504, in _call_chain
    result = func(*args)
  File "C:\Users\…\Python\Python36-32\lib\urllib\request.py", line 650, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 404: Not Found

How can I solve it?