Undefined variable defined within try-except block

Question

I am using the below code to scrape over XFN content from web page http://ajaxian.com but I am getting undefined variable error:

My code is as follows:

'''
Created on Jan 11, 2013

@author: Somnath
'''
# Scraping XFN content from a web page
# -*-coding: utf-8 -*-

import sys
import urllib2
import HTMLParser
from BeautifulSoup import BeautifulSoup

# Try http://ajaxian.com
URL = sys.argv[0]

XFN_TAGS = set([
            'colleague',
            'sweetheart',
            'parent',
            'co-resident',
            'co-worker',
            'muse',
            'neighbor',
            'sibling',
            'kin',
            'child',
            'date',
            'spouse',
            'me',
            'acquaintance',
            'met',
            'crush',
            'contact',
            'friend',
            ])


try:
    page = urllib2.urlopen(URL)
except urllib2.URLError:
    print 'Failed to fetch ' + item

try:
    soup = BeautifulSoup(page)
except HTMLParser.HTMLParseError:
    print 'Failed to parse ' + item

anchorTags = soup.findAll('a')

for a in anchorTags:
    if a.has_key('rel'):
        if len(set(a['rel'].split()) & XFN_TAGS) > 0:
            tags = a['rel'].split()
            print a.contents[0], a['href'], tags

I have two try blocks in my code and it is giving an error undefined variable : item. If I want to re-include the try-except blocks, should I give a blank definition of variable, item outside the try blocks?

PS: Please note that is a standard code followed from a book. And I expect that they would not have made such a trivial mistake. Am I getting something wrong here ?

Answer 1

Assuming that you want to print the URL that failed to load, try changing it to print 'Failed to fetch ' + URL . You aren't actually defining item anywhere, so Python doesn't know what you mean:

try:
    page = urllib2.urlopen(URL)
except urllib2.URLError:
    print 'Failed to fetch ' + URL

And in your second block, change item to URL as well (assuming the error you want to display shows the URL and not the content).

try:
    soup = BeautifulSoup(page)
except HTMLParser.HTMLParseError:
    print 'Failed to parse ' + URL

Answer 2

print 'Failed to fetch ' + item

item is not defined any where. I guess you wanted to print URL there.

As per python tutorial

Variables must be “defined” (assigned a value) before they can be used, or an error will occur:

Answer 3

You did not define the variable 'item'. That's what is causing the error. You must define a variable before you use it.