Bane — Sat, 08 Sep 2001 10:27:16 +0000

Thank you so much for your help! I am in your debt

Mark was nice enough to assist me and I'm happy to say my problem is not only solved but solved beyond my expectations Thank you both for your assisstance.

Mark Hensler — Thu, 06 Sep 2001 01:30:00 +0000

Welcome to TWF, Bane!

I would use an array...

<?php
$dont_log = array(
    \"bob.com\",with bob.com
    \"bill.com\",
    \"spankysue.com\"
    );

if (!in_array($HTTP_REFERER, $dont_log) && !in_array(\"www.$HTTP_REFERER\",$dont_log)) {
    //domain not found, log it
    
}
?>

You could even get more creative, and use regex..

<?php
$dont_log = array(
    \"/([0-9a-zA-Z]*)bob.com$/i\"
    \"/^([w\\.]*)bill.com$/\",
    \"/spankysue.com/\"
    );

foreach ($dont_log as $tmp) {
    if (!preg_match($tmp,$HTTP_REFERER)) {
        //domain not found, log it
        
    }
}
?>

Docs: preg_match(), in_array()

tmay — Wed, 05 Sep 2001 21:49:50 +0000

No problem. I don't know PHP though, sorry.

Good Luck!

Bane — Wed, 05 Sep 2001 21:44:51 +0000

I suppose I should have said with PHP Thanks for the assistance however

tmay — Wed, 05 Sep 2001 21:42:07 +0000

I'm assuming this is Perl? If you only need to search through it once, I'd stick them into an array and sift through them with a foreach statement:

@skip = qw/bob.com bill.com spankysue.com/;

$is_there = 0;
    foreach $i (@skip) {
	if ($i eq $domain_to_find) {
	    $is_there = 1;
	    last;
	}
    }
    if ($is_there) { ...code not to log... }