Replace url with link using regex in python

Question

how to convert text to link? Back in PHP, I used this piece of code that worked well for my purpose:

            $text = preg_replace("#(^|[\n ])(([\w]+?://[\w\#$%&~.\-;:=,?@\[\]+]*)(/[\w\#$%&~/.\-;:=,?@\[\]+]*)?)#is", "\\1<a href=\"\\2\" target=\"_blank\">\\3</a>", $text);
            $text = preg_replace("#(^|[\n ])(((www|ftp)\.[\w\#$%&~.\-;:=,?@\[\]+]*)(/[\w\#$%&~/.\-;:=,?@\[\]+]*)?)#is", "\\1<a href=\"http://\\2\" target=\"_blank\">\\3</a>", $text);

I tried in Python but could not get it to work. It would be very nice if someone could translate this into Python :) ..

Answer 1

The code below is a simple python translation. You must confirm that he really does what you want. See the POWON Regular Expression HOWTO for more information .

import re

pat1 = re.compile(r"(^|[\n ])(([\w]+?://[\w\#$%&~.\-;:=,?@\[\]+]*)(/[\w\#$%&~/.\-;:=,?@\[\]+]*)?)", re.IGNORECASE | re.DOTALL)

pat2 = re.compile(r"#(^|[\n ])(((www|ftp)\.[\w\#$%&~.\-;:=,?@\[\]+]*)(/[\w\#$%&~/.\-;:=,?@\[\]+]*)?)", re.IGNORECASE | re.DOTALL)


urlstr = 'http://www.example.com/foo/bar.html'

urlstr = pat1.sub(r'\1<a href="\2" target="_blank">\3</a>', urlstr)
urlstr = pat2.sub(r'\1<a href="http:/\2" target="_blank">\3</a>', urlstr)

print urlstr

Here is the result at my end:

<a href="http://www.example.com/foo/bar.html" target="_blank">http://www.example.com</a>