Removing top-directory-only URLs from a list of URLs? - perl

Question

I have a question that I'm having trouble researching, as I don't know how to ask it correctly on a search engine.
I have a list of URLs. I would like to have some automated way (Perl for preference) to go through the list and remove all URLs that are top directory only.
So for example I might have this list:
http://www.example.com/hello.html
http://www.foo.com/this/thingrighthere.html
In this case I would want to remove example.com from my list, as it is either top-directory only or they reference files in a top directory.
I'm trying to figure out how to do that. My first thought was, count forward slashes and if there's more than two, eliminate the URL from the list. But then you have trailing forward slashes, so that wouldn't work.
Any ideas or thoughts would be much appreciated.

Answer 1

Something like this:
use URI::Split qw( uri_split );
my $url = "http://www.foo.com/this/thingrighthere.html";
my ($scheme, $auth, $path, $query, $frag) = uri_split( $url );
if (($path =~ tr/\///) > 1 ) {
print "I care about this $url";
}
http://metacpan.org/pod/URI::Split

Answer 2

You could do this with regexes, but its much less work to let the URI library do it for you. You won't get caught out by funny schemes, escapes, and extra stuff before and after the path (query, anchor, authorization...). There's some trickiness around how paths are represented by path_segments(). See the comments below and the URI docs for details.
I have assumed that http://www.example.com/foo/ is considered a top directory. Adjust as necessary, but its something you have to think about.
#!/usr/bin/env perl
use URI;
use File::Spec;
use strict;
use warnings;
use Test::More 'no_plan';
sub is_top_level_uri {
my $uri = shift;
# turn it into a URI object if it isn't already
$uri = URI->new($uri) unless eval { $uri->isa("URI") };
# normalize it
$uri = $uri->canonical;
# split the path part into pieces
my #path_segments = $uri->path_segments;
# for an absolute path, which most are, the absoluteness will be
# represented by an empty string. Also /foo/ will come out as two elements.
# Strip that all out, it gets in our way for this purpose.
#path_segments = grep { $_ ne '' } #path_segments;
return #path_segments <= 1;
}
my #filtered_uris = (
"http://www.example.com/hello.html",
"http://www.example.com/",
"http://www.example.com",
"https://www.example.com/",
"https://www.example.com/foo/#extra",
"ftp://www.example.com/foo",
"ftp://www.example.com/foo/",
"https://www.example.com/foo/#extra",
"https://www.example.com/foo/?extra",
"http://www.example.com/hello.html#extra",
"http://www.example.com/hello.html?extra",
"file:///foo",
"file:///foo/",
"file:///foo.txt",
);
my #unfiltered_uris = (
"http://www.foo.com/this/thingrighthere.html",
"https://www.example.com/foo/bar",
"ftp://www.example.com/foo/bar/",
"file:///foo/bar",
"file:///foo/bar.txt",
);
for my $uri (#filtered_uris) {
ok is_top_level_uri($uri), $uri;
}
for my $uri (#unfiltered_uris) {
ok !is_top_level_uri($uri), $uri;
}

Answer 3

Use the URI module from CPAN. http://search.cpan.org/dist/URI
This is a solved problem. People have already written, tested and debugged code that handles this already. Whenever you have a programming problem that others have probably had to deal with, then look for existing code that does it for you.