closing files properly opened with urllib2.urlopen()

Question

I have following code in a python script

  try:
    # send the query request
    sf = urllib2.urlopen(search_query)
    search_soup = BeautifulSoup.BeautifulStoneSoup(sf.read())
    sf.close()
  except Exception, err:
    print("Couldn't get programme information.")
    print(str(err))
    return

I'm concerned because if I encounter an error on sf.read(), then sf.clsoe() is not called. I tried putting sf.close() in a finally block, but if there's an exception on urlopen() then there's no file to close and I encounter an exception in the finally block!

So then I tried

  try:
    with urllib2.urlopen(search_query) as sf:
      search_soup = BeautifulSoup.BeautifulStoneSoup(sf.read())
  except Exception, err:
    print("Couldn't get programme information.")
    print(str(err))
    return

but this raised a invalid syntax error on the with... line. How can I best handle this, I feel stupid!

As commenters have pointed out, I am using Pys60 which is python 2.5.4

Answer 1

I would use contextlib.closing (in combination with from __future__ import with_statement for old Python versions):

from contextlib import closing

with closing(urllib2.urlopen('http://blah')) as sf:
    search_soup = BeautifulSoup.BeautifulStoneSoup(sf.read())

Or, if you want to avoid the with statement:

try:
    sf = None
    sf = urllib2.urlopen('http://blah')
    search_soup = BeautifulSoup.BeautifulStoneSoup(sf.read())
finally:
    if sf:
        sf.close()

Not quite as elegant though.

Answer 2

finally:
    if sf: sf.close()