Python http error logging

Question

following question

I do have a python script working and doing HTTP requests. Sometimes I get several errors like the http 400. Now I want to log those and interrupt the script from stop working.

 import logging

 date = .strftime().....

 try: 
        script X
 except:
        error = urllib2.HTTPError
        errorcode = error.code
        file = open ("error.txt",'r+')
        file.write("Error Code: %s, date \n", errorcode)
        file.close()
        pass

Is that right? Or did I something wrong. Should I use an If for the HTTP error?

Will this prevent the script from stopping while getting an error -> pass

Thank You!

Why cant you just try it yourself?

Vishal
– Vishal

2015-08-11 08:08:45 +00:00
Commented Aug 11, 2015 at 8:08 — Vishal
– Vishal, Commented Aug 11, 2015 at 8:08

Open AI - Opting Out · Accepted Answer · 2015-08-11 12:11:47Z

1

Your line saying:

error = urllib2.HTTPError

makes no sense in this context. It is saying error is the urllib2.HTTPError class. That class is used for raising exceptions, you are interested in catching exceptions.

Usually we write:

try: 
    #  http attempt
except urllib2.HTTPError, error:
    errorcode = error.code

This says the type of the exception you are handling, urllib2.HTTPError, and that is what type error will be, or any class object which inherits from urllib2.HTTPError.

It also means you won't handle any other kinds of exceptions. This is a good thing. Only deal with what you can handle and let everything else fly past you.

At the moment your except statement catches every possible exception type. You also throw away the information, so you don't really know what the exception was.

Writing to a file, use with. This makes sure the file is closed properly if there is an exception whilst writing to it. Don't use file as a variable name. You need to interpolate the string format using %:

    with open ("error.txt",'r+') as log:
        log.write("Error Code: %s, date \n" % errorcode)

Okay, code review over.

edited Aug 11, 2015 at 12:11

answered Aug 11, 2015 at 9:03

Open AI - Opting Out

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2 Comments

Ser1 Over a year ago

oh right, thank you, my bad. So I will log the error but still use 'pass' for 'going on'

Open AI - Opting Out Over a year ago

You only need pass if the block would otherwise be empty. See this question.

score 1 · Accepted Answer · 2015-08-11 08:18:38Z

1

Use python requests library::::

try:
    res = requests.post(url, data, headers=headers, auth=auth)
    if res.status_code == 200 or res.status_code == 201:
        return 'SUCCESS', res.content
    else:
        return 'FAIL', res.status_code
except Exception as e:
    print "Http request failed... %s" % traceback.format_exc()

edited Aug 11, 2015 at 8:18

answered Aug 11, 2015 at 8:08

user2879704

3 Comments

Burhan Khalid Over a year ago

You should use raise_for_status.

Ser1 Over a year ago

but your method is w/o a error logging. Only printing the error isn't my aim. Furthermore I used the urllib2 instead of request. Should I change it to the request method?

Open AI - Opting Out Over a year ago

Don't recommend installing a library for a basic question.

Burhan Khalid · Accepted Answer · 2015-08-11 08:25:24Z

0

A better way to structure your code is to use error logging (a feature built into Python):

import logging

logger = logging.getLogger('__name__')

fh = logging.FileHandler('errors.log')
logger.addHandler(fh)

try:
   # something
except urllib2.HttpError, e:
   logger.error(e)

answered Aug 11, 2015 at 8:25

Burhan Khalid

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