Python howto exception handle <urlopen error [Errno 54] Connection reset by peer>

Question

My script in Python 2.7 scrapes a website every minute, but sometimes it gives the error:

urlopen error [Errno 54] Connection reset by peer>

How I can handle the exception? I am trying something like this:

from socket import error as SocketError
import errno

try:
    response = urllib2.urlopen(request).read()
except SocketError as e:
    if e.errno != errno.ECONNRESET:
        print "there is a time-out" 
    pass
        print "There is no time-out, continue"

Asif Raza · Accepted Answer · 2017-06-15 12:55:06Z

2

You can handle the exception like this:

from socket import error as SocketError
import errno    

try:
    urllib2.urlopen(request).read()
except SocketError, e:
     errorcode = e[0]
     if errorcode!=errno.ECONNREFUSED:
        print "There is a time-out"
        # Not the error we are looking for, re-raise
        raise e

You can read Error Code

edited Jun 15, 2017 at 12:55

answered Jun 15, 2017 at 9:21

Asif Raza

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7 Comments

Jesse kraal Over a year ago

I have import your code into my script and after a half hour I receive the error: except SocketError, e: NameError: name 'SocketError' is not defined.

Asif Raza Over a year ago

Add this line from socket import error as SocketError above errno.

Jesse kraal Over a year ago

I replaced except SocketError, e: in except socket.error as e: I will try this out if it works. Thanks for your help

Jesse kraal Over a year ago

Oh I forgot your line of code! Ok I will try it again.

Asif Raza Over a year ago

If still your problem not solved try this Question

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