用urllib2.urlopen()正确打开关闭文件

Question

我在python脚本中有以下代码

  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

我很担心,因为如果我遇到错误sf.read(),那么sf.clsoe()就不会被调用.我尝试放入sf.close()一个finally块,但是如果有一个异常,urlopen()则没有文件要关闭,我在finally块中遇到异常!

所以我试过了

  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

但这会with...在行上引发无效的语法错误.我怎么能最好地处理这个问题,我觉得很蠢!

正如评论者指出的那样,我使用的是Pys60,它是python 2.5.4

Answer 1

我会使用contextlib.closing(与旧的Python版本的__future__ import with_statement结合使用):

from contextlib import closing

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

或者,如果您想避免使用with语句:

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

虽然不那么优雅.

Answer 2

finally:
    if sf: sf.close()

Answer 3

为什么不尝试关闭sf,如果它不存在则传递？

import urllib2
try:
    search_query = 'http://blah'
    sf = urllib2.urlopen(search_query)
    search_soup = BeautifulSoup.BeautifulStoneSoup(sf.read())
except urllib2.URLError, err:
    print(err.reason)
finally:
    try:
        sf.close()
    except NameError: 
        pass