Question

so what I want to do is remove everything after and including the first “/” to appear after a “.”
so: http://linux.pacific.net.au/primary.xml.gz
would become: http://linux.pacific.net.au

How do I do this using regex? The system I’m running on can’t use URI tool.

Answer 1

I suggest you use URI::Split which will separate a standard URL into its constuent parts for you and rejoin them. You want the first two parts – the scheme and the host.

use strict;
use warnings;

use URI::Split qw/ uri_split uri_join /;

my $scheme_host = do {
  my (@parts) = uri_split 'http://linux.pacific.net.au/primary.xml.gz';
  uri_join @parts[0,1];
};

print $scheme_host;

output

http://linux.pacific.net.au

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Update

If your comment The system I’m running on can’t use URI tool means you can’t install modules, then here is a regular expression solution.

You say you want to remove everything after and including the first “/” to appear after a “.”, so /^.*?\./ finds the first dot, and m|[^/]+| finds everything after it up tot he next slash.

The output is identical to that of the preceding code

use strict;
use warnings;

my $url = 'http://linux.pacific.net.au/primary.xml.gz';

my ($scheme_host) = $url =~ m|^( .*?\. [^/]+ )|x;

print $scheme_host;