Parsing url inside string

How would I go about matching the

following string format ( everything after the equal sign to the end of the .html

http%3A%2F%2Fwww.mydomains.com.com%2FSA100.html

i开发者_StackOverflow社区nside the string below:

http://www.tticker.com/me0439-119?url=http%3A%2F%2Fwww.mydomains.com.com%2FSA100.html%3Fc%2acn%2CSA400

(?<==).*?\.html

test with grep

   kent$  echo "http://www.tticker.com/me0439-119?url=http%3A%2F%2Fwww.mydomains.com.com%2FSA100.html%3Fc%2acn%2CSA400"|grep -Po "(?<==).*?\.html"
    http%3A%2F%2Fwww.mydomains.com.com%2FSA100.html

Simplest I could come up with was:

/url=(http.*\.html)/

Use the capture group for your URL.

In perl:

#!/usr/bin/perl -w
use URI;
my $uri = URI->new("http://www.tticker.com/me0439-119?url=http%3A%2F%2Fwww.mydomains.com.com%2FSA100.html%3Fc%2acn%2CSA400"); # create URI object
my %params = $uri->query_form();                      # get all params
my $param_url = $params{url};

my $uri2 = URI->new($param_url); # create new URI object from param URL  
$uri2->query(undef);             # strip parameters
print $uri2->as_string();

gives:

http://www.mydomains.com.com/SA100.html