[SOLVED] Exception handling in Python

Turbocapitalist · 04-26-2020, 08:06 AM

I'm wondering how to trap exceptions in Python. Specifically, when I run this code:

Code:

#!/usr/bin/python3                                                              

import urllib.request

uri = 'http://localhost:1880/'
req = urllib.request.Request(uri,method="HEAD",data=None)

with urllib.request.urlopen(uri) as response:
    try:
        res = response.read()
    except HTTPError as e:
        print("HTTP error")
    except URLError as e:
        print("URL error")
    except TypeError as e:
        print("Type error")
    except ValueError as e:
        print("Value error")

and try to connect it to a session which will abort,

Code:

timeout 2 nc -l 1880 & ./script.py

then I always get the following crash report:

Code:

GET / HTTP/1.1
Accept-Encoding: identity
Host: localhost:1880
User-Agent: Python-urllib/3.7
Connection: close

Traceback (most recent call last):
  File "./script.py", line 10, in <module>                  
    with urllib.request.urlopen(uri) as response:                              
  File "/usr/lib/python3.7/urllib/request.py", line 222, in urlopen            
    return opener.open(url, data, timeout)                                     
  File "/usr/lib/python3.7/urllib/request.py", line 525, in open               
    response = self._open(req, data)
  File "/usr/lib/python3.7/urllib/request.py", line 543, in _open              
    '_open', req)
  File "/usr/lib/python3.7/urllib/request.py", line 503, in _call_chain        
    result = func(*args)
  File "/usr/lib/python3.7/urllib/request.py", line 1345, in http_open         
    return self.do_open(http.client.HTTPConnection, req)                       
  File "/usr/lib/python3.7/urllib/request.py", line 1320, in do_open           
    r = h.getresponse()
  File "/usr/lib/python3.7/http/client.py", line 1336, in getresponse          
    response.begin()
  File "/usr/lib/python3.7/http/client.py", line 306, in begin                 
    version, status, reason = self._read_status()                              
  File "/usr/lib/python3.7/http/client.py", line 275, in _read_status          
    raise RemoteDisconnected("Remote end closed connection without"            
http.client.RemoteDisconnected: Remote end closed connection without response

What is the right way to catch that error so the script can make note of it and continue? Right now the script just dies.

dugan · 04-26-2020, 10:07 AM

Code:

import http

# ...

except http.client.RemoteDisconnected as e:

From what I'm seeing of the script though, you should be using a ready-made solution like Postman or HTTPie.

Turbocapitalist · 04-26-2020, 11:03 AM

Quote:

Originally Posted by dugan

Code:

import http

# ...

except http.client.RemoteDisconnected as e:

Hmm. I get the same result even adding those into the script above.

SoftSprocket · 04-26-2020, 11:42 AM

If you want to cacth the exception in urllib.request.urlopen you need to move it into the scope of the try. As is it's outside so not caught.

dugan · 04-26-2020, 12:28 PM

Quote:

Originally Posted by Turbocapitalist

Hmm. I get the same result even adding those into the script above.

Then post your revised script.

Turbocapitalist · 04-26-2020, 12:31 PM

Here is how I have tried it:

Code:

#!/usr/bin/python3                                                              

import urllib.request
import http

uri = 'http://localhost:1880/'
req = urllib.request.Request(uri,method="HEAD",data=None)

with urllib.request.urlopen(uri) as response:
    try:
        res = response.read()
    except http.client.RemoteDisconnected as e:
        print("Disconnect")
    except HTTPError as e:
        print("HTTP error")
    except URLError as e:
        print("URL error")
    except TypeError as e:
        print("Type error")
    except ValueError as e:
        print("Value error")

I have only a few tens of hours of experience with Python.

teckk · 04-26-2020, 12:37 PM

Code:

#!/usr/bin/python

from urllib import request
from urllib.error import URLError
        
agent = ('Mozilla/5.0 (Windows NT 10.1; x86_64; rv:75.0)'
        ' Gecko/20100101 Firefox/75.0')
                
user_agent = {'User-Agent': agent}
        
uri1 = 'https://www.google.com'
uri2 = 'https://1234abc.com'

try:
    req = request.Request(uri1, data=None,  headers=user_agent)
    page = request.urlopen(req)
    res = page.read()
    print('uri1 is OK')

except URLError as e:
    print('uri1 is URLError')
    
try:
    req = request.Request(uri2, data=None,  headers=user_agent)
    page = request.urlopen(req)
    res = page.read()
    print('uri2 is OK')

except URLError as e:
    print('uri2 is URLError')

Turbocapitalist · 04-26-2020, 12:50 PM

Oops. How embarrassing. It was both the spacing and the sequence of 'try'

Code:

. . .
try:
    with urllib.request.urlopen(uri) as response:
        res = response.read()
. . .

So far, I think I've spent about 75% of my time with the white space in regards to python.

teckk · 04-26-2020, 01:09 PM

So I think that would be something like:

test1.py

Code:

#!/usr/bin/python

from urllib import request
from urllib.error import HTTPError, URLError

uri = 'http://127.0.0.1:1880/'
req = request.Request(uri)

def Except():
    try:
        page = request.urlopen(req)
        res = page.read()
        print('It is ok')
        
    except HTTPError as e:
        print("HTTP error")
        
    except URLError as e:
        print("URL error")
        
    except TypeError as e:
        print("Type error")
        
    except ValueError as e:
        print("Value error")
        
if __name__ == "__main__":
    Except()

Test it.

In one terminal

Code:

python3 -m http.server 1880

In second terminal

Code:

python ./test1.py

Turbocapitalist · 04-26-2020, 01:11 PM

Thanks. I'm now digesting the changes and the suggesstions, and trying various tests.