Why does my urllib.request return a http error 403?

Question

i am attempting to make a program that downloads a series of product pictures from a site using python. The site stores its images under a certain url format https://www.sitename.com/XYZabcde where XYZ are three letters that represent the brand of the product and abcde are a series of numbers in between 00000 and 30000. here is my code:

import urllib.request

def down(i, inp):
    full_path = 'images/image-{}.jpg'.format(i)
    url = "https://www.sitename.com/{}{}.jpg".format(inp,i)
    urllib.request.urlretrieve(url, full_path)

    print("saved")
    return None

inp = input("brand :" )

i = 20100

while i <= 20105:
    x = str(i)
    y = x.zfill(5)
    z = "https://www.sitename.com/{}{}.jpg".format(inp,y)
    print(z)
    down(y, inp)
    i += 1

With the code i have written i can successfully download a series of pictures from it which i know exist for example brand RVL from 20100 to 20105 will succesfully download those six pictures. however when i broaden the while loop to include links i dont know will give me an image i get this error code :

Traceback (most recent call last):
  File "c:/Users/euan/Desktop/university/programming/Python/parser/test - Copy.py", line 20, in <module>
    down(y, inp)
  File "c:/Users/euan/Desktop/university/programming/Python/parser/test - Copy.py", line 6, in down
    urllib.request.urlretrieve(url, full_path)
  File "C:\Users\euan\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 247, in urlretrieve
    with contextlib.closing(urlopen(url, data)) as fp:
  File "C:\Users\euan\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 222, in urlopen
    return opener.open(url, data, timeout)
  File "C:\Users\euan\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 531, in open
    response = meth(req, response)
  File "C:\Users\euan\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 640, in http_response
    response = self.parent.error(
  File "C:\Users\euan\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 569, in error
    return self._call_chain(*args)
  File "C:\Users\euan\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 502, in _call_chain
    result = func(*args)
  File "C:\Users\euan\AppData\Local\Programs\Python\Python38\lib\urllib\request.py", line 649, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden

what can i do to check and avoid any url that would yield this result?

Answer 1

You cannot as such know in advance which URLs you don't have access to, but you can surround the download with a try-except:

import urllib.request, urllib.error

...

def down(i, inp):
    full_path = 'images/image-{}.jpg'.format(i)
    url = "https://www.sitename.com/{}{}.jpg".format(inp,i)
    try:
        urllib.request.urlretrieve(url, full_path)
        print("saved")
    except urllib.error.HTTPError as e:
        print("failed:", e)


    return None

In that case it will just print eg "failed: HTTP Error 403: Forbidden" whenever a URL cannot be fetched, and the program will continue.