Question

I have copied this code from the other SO question but i am not able to solve the error i am getting

this is my code

import os
import urllib2
import sys
d = urllib2.urlopen("http://www.google.com.au/logos/2012/new_years_eve_2012_-_english_only-1049005-hp.jpg")
o = open('image.%s' % d.info().gettype(), 'w')
o.write(d.read())

The error is

Traceback (most recent call last): File “./image_test.py”, line 6,
in
o = open(‘image.%s’ % d.info().gettype(), ‘w’) IOError: [Errno 2] No such file or directory: ‘image.image/jpeg’

Answer 1

This is because HTML MIME-types typically include slash, e.g. image/jpeg. That causes confusion for open, since slash is a special character used in file system routing.

You just have to avoid slash in a file name, for example in a way Joran proposed.

UPDATE In order to make it work in all situations, you have to get a map MIME-type -> file extension. There’s one at Apache: http://svn.apache.org/viewvc/httpd/httpd/trunk/docs/conf/mime.types?view=markup . Once you parse it to python dictionary like this:

import re

mimetypes = {}
with open('mime.types') as mimedata:
    for line in [line.strip() for line in mimedata.readlines()]:
        if line[0] == '#':
            continue

        # This pattern will only get the first extension in the list
        pattern = re.compile(r'([^\s]+)\s+([^\s]+)')
        result  = pattern.match(line)
        mime    = result.group(1)
        ext     = result.group(2)
        mimetypes[mime] = ext

you will use it the following way:

mime = d.info().gettype()
ext = ''
if mime in mimetypes:
    ext = mimetypes[mime]

o = open('data' + ext, 'w')
o.write(d.read())