getting the print version

[eremiya]

How to get print version if link to page view as:
http://domain.tld/category/subcategory/[number]-[title]
and print version:
http://domain.tld/category/subcategory/[number]/print

Actually I need to replace the "hyphen" and everything that comes after it to "/print"

[Bobus]

I think:

import re

def print_version(self, url):
    return re.sub(r'([0-9]*)\-.*', r'\g<1>/print', url)

if it's not work, send the right url.

[eremiya]

Uunfortunately did not work.

XML: http://k.img.com.ua/rss/ru/kyiv.xml
Example page: http://korrespondent.net/kyiv/119700...ka-v-etom-godu
Example print page: http://korrespondent.net/kyiv/1197007/print

[Bobus]

    def print_version(self, url):
        return url+'/print'

Because in the feed only the number is given.