Http message is not callable porting python2 to python3 download script - python-3.x

Question

i'am trying to port to python 3 this script :
import re
from os.path import basename
import os
from urllib.parse import urlparse,urlsplit
from urllib.request import urlopen,Request
import urllib
def url2name(url):
return basename(urlsplit(url)[2])
def download(url, out_path="."):
localName = url2name(url)
req = Request(url)
r = urlopen(req)
if r.info().has_key('Content-Disposition'):
# If the response has Content-Disposition, we take file name from it
localName = r.info()['Content-Disposition'].split('filename=')[1]
if localName[0] == '"' or localName[0] == "'":
localName = localName[1:-1]
elif r.url != url:
# if we were redirected, the real file name we take from the final URL
localName = url2name(r.url)
localName = os.path.join(out_path, localName)
f = open(localName, 'wb')
f.write(r.read())
f.close()
but i have a :
'HTTPMessage' object is not callable
r.info() seems to have problems
how to get the header info in python 3 ?

Answer 1

Try with this, you should use context managers:
def download(url, out_path="."):
localName = url2name(url)
req = Request(url)
with urlopen(req) as f:
content_disp = f.getheader('Content-Disposition')
if content_disp:
# If the response has Content-Disposition, we take file name from it
localName = content_disp.split('filename=')[1]
if localName[0] == '"' or localName[0] == "'":
localName = localName[1:-1]
elif f.url != url:
# if we were redirected, the real file name we take from the final URL
localName = url2name(f.url)
localName = os.path.join(out_path, localName)
with open(localName, 'wb') as fp:
fp.write(f.read())