CODEWITHSUNDEEP
pythonpython-3.xpython-2.7urllib
probably-asskicker
probably-asskicker

Reputation: 171

URLLIB request code reading issue. Cannot read error code 404

I am trying to read the url code using URLLIB. Here is my code:

import urllib

url = "https://www.facebook.com/fads0000fass"

r = urllib.request.urlopen(url)
p = r.code

if(p == "HTTP Error 404: Not Found" ):
        print("hello")
else:
        print("null")

The url I am using will show error code 404 but I am not able to read it. I also tried if(p == 404) but I get the same issue. I Can read other codes i.e. 200, 201 etc.

Can you please help me fix it?

traceback:

Traceback (most recent call last):
  File "gd.py", line 7, in <module>
    r = urllib.request.urlopen(url)
  File "/usr/lib64/python3.7/urllib/request.py", line 222, in urlopen
    return opener.open(url, data, timeout)
  File "/usr/lib64/python3.7/urllib/request.py", line 531, in open
    response = meth(req, response)
  File "/usr/lib64/python3.7/urllib/request.py", line 641, in http_response
    'http', request, response, code, msg, hdrs)
  File "/usr/lib64/python3.7/urllib/request.py", line 569, in error
    return self._call_chain(*args)
  File "/usr/lib64/python3.7/urllib/request.py", line 503, in _call_chain
    result = func(*args)
  File "/usr/lib64/python3.7/urllib/request.py", line 649, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 404: Not Found

Upvotes: 1

Views: 3217

Answers (2)

Sam Morgan
Sam Morgan

Reputation: 3318

In order to reach your if statement, your code needs exception handling. An exception is being raised when you call urlopen on line 7. See the first step of your traceback.

File "gd.py", line 7, in <module>
  r = urllib.request.urlopen(url)

The exception happens here, which causes your code to exit, so further statements aren't evaluated. To get past this, you must handle the exception.

import urllib.request

url = "https://www.facebook.com/fads0000fass"

try:
    r = urllib.request.urlopen(url)
except urllib.error.HTTPError as e:
    # More useful
    # print(f"{e.code}: {e.reason}\n\n{e.headers}")
    if e.code in [404]:
        print("hello")
    else:
        print("null")

Going beyond this, if you want something more like your original logic, I'd recommend using the requests library. I'd actually recommend using requests for all of your HTTP needs whenever possible, it's exceptionally good.

import requests

r = requests.get(url)
p = r.status_code

if r.status_code == 404:
    print("hello")
else:
    print("null")

Upvotes: 0

antho
antho

Reputation: 638

I'm not sure that's what you're asking. 

import urllib.request

url = "https://www.facebook.com/fads0000fass"

try:
    r = urllib.request.urlopen(url)
    p = r.code
except urllib.error.HTTPError:
    print("hello")

Upvotes: 1

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