getting the file size with pycurl

Question

I want to write a downloader with python and I use PycURL as my library, but I got a problem. I can't get the size of the file wich I wanna download. Here is part of my code :

import pycurl
url = 'http://www.google.com'
c = pycurl.Curl()
c.setopt(c.URL, url)
print c.getinfo(c.CONTENT_LENGTH_DOWNLOAD)
c.perform()

When I test this code in python shell, it's ok but when I write it as a function and run it, it gives me -1 instead of the size. What is the problem?

(code's been edited)

Answer 1

From the pycurl documentation on the Curl object:

The getinfo method should not be called unless perform has been called and finished.

You're calling getinfo before you've called perform.

Here is a simplified version of your example, does this work?

import pycurl

url = 'http://www.google.com'
c = pycurl.Curl()
c.setopt(c.URL, url)
c.perform()
print c.getinfo(c.CONTENT_LENGTH_DOWNLOAD)

You should see the HTML content followed by the size.

Answer 2

This answer adds the missing c.setopt(c.NOBODY, 1) and is otherwise the same as the one given some months ago:

import pycurl

c = pycurl.Curl()
c.setopt(c.URL, 'http://www.alfe.de')
c.setopt(c.NOBODY, 1)
c.perform()
c.getinfo(c.CONTENT_LENGTH_DOWNLOAD)

Calling c.setopt(c.NOBODY, 1) before calling c.perform() avoids downloading the contents of the file ("No Body", but all headers).

Answer 3

Try adding debug to see what happens actually. After you created curl make this:

def curl_debug(debug_type, msg):
    print("debug: %s %s" % (repr(debug_type), repr(msg)))

c.setopt(pycurl.VERBOSE, 1)
c.setopt(pycurl.DEBUGFUNCTION, curl_debug)