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Getting the URL of a referer

by ACJavascript (Acolyte)
on Dec 23, 2003 at 11:49 UTC ( [id://316652]=perlquestion: print w/replies, xml ) Need Help??

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ACJavascript has asked for the wisdom of the Perl Monks concerning the following question:

Hello all,

I SEEK THE WISDOM OF THE PERL MONKS! hehe

Okay I need to know this:
Say the referer is http://www.yahoo.com/s=2342&stuff=asdfsd

I read the referer using ENV{'HTTP_REFERER'} - NOW
How can I split it up so that All I have like is www.yahoo.com?

Im pretty sure its going somthing to do with substr but I can't figure it out. Please any help would be most appreciated!
A*C

Replies are listed 'Best First'.
Re: Getting the URL of a referer
by barrd (Canon) on Dec 23, 2003 at 12:37 UTC
    Rather than "do it for you", have a look at this link... then you'll be able to do it yourself.

    Much more satisfying that way. ;)

Re: Getting the URL of a referer
by chromatic (Archbishop) on Dec 23, 2003 at 13:13 UTC

    Alternately, use the URI module and the host() method.

Re: Getting the URL of a referer
by Thelonius (Priest) on Dec 23, 2003 at 13:18 UTC
    First, this "http://www.yahoo.com/s=2342&stuff=asdfsd" is a URL; this "www.yahoo.com" is a domain name or host name. Second, be aware that the referer is not always present.

    There are many ways to do this (say, maybe that should be a slogan :-).

    The best way, if you have the URI module installed, is this:

    use URI::URL; my $host = URI::URL->new($ENV{'HTTP_REFERER'})->host;
    That will handle more obscure cases, such as "https://joe@www.yahoo.com:4040/fred?billy".
      There are many ways to do this (say, maybe that should be a slogan :-).

      Sorry man, you're a little too late for that. We've already got the slogan for that. Yours is so close I can't help but wonder if you already know it and the smiley emoticon is hinting at the slogan. Here t'is:

      TIMTOWTDI: There Is More Than One Way To Do It

        Thanks alot guys/gals I will try all of your suggestions!!!

        A*C

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