[Resolved] Current Page Url Variable???

[dkin69]

Current Page Url Variable???

Ok this is my problem.

I have a script I am using as an include in many sites. It pulls info from a database and displays ads. Now every site that this is displayed on has an ad in the database and I dont want anyones ads showing on their own site.

I wrote this.

##$current = $_SERVER["HTTP_HOST"];
$current = $_SERVER['PHP_SELF'];

$lin = mysql_query("SELECT * FROM link order by id asc", $link) or die ("query 1: " . mysql_error());
echo '<li><h2>LinkWork</h2>'
. '<ul>';
while ($li = mysql_fetch_array($lin)) {
if ($current == $li['base'])
{
echo '';
}
else
{
echo '<li><a href="'.$li['web_url'].'" alt="'.$li['alt_tag'].'">'.$li['display'].'</a></li>';

}
}

The column base, is the base of a url eg. digitalpoint.com

So that is what I need to find, a php variable that gives me the current pages web address, but only up to the end of the com or org.

 

[-NC-]

Something like this?

$domain = str_replace("www." ,"", $_SERVER['SERVER_NAME']);

$lin = mysql_query("SELECT * FROM link WHERE base = '" .$domain ."' order by id asc", $link)

 

[mikesherov]

$url =  $_SERVER['SCRIPT_URI'];
$parsed_url = parse_url($url);
$domain = $parsed_url['hostname'];

at this point, $domain should be whatever the domain portion is.

To see more information about the PHP parse_url() function, go here: http://us2.php.net/parse_url

 

[dkin69]

ok, I had to alter my code, it is now this

Include.php

$lin = mysql_query("SELECT * FROM link order by id asc", $link) or die ("query 1: " . mysql_error());
echo '<li><h2>LinkWork</h2>'
. '<ul>';
while ($li = mysql_fetch_array($lin)) {
if ($current == $li['base'])
{
echo '';
}
else
{
echo '<li><a href="'.$li['web_url'].'" alt="'.$li['alt_tag'].'">'.$li['display'].'</a></li>';

}
}

Including file

<?php
$current = 'bluewidget';
include("include.php");?>

This is on a wordpress blog if that helps.

and the variable in the including file just doesn't seem to be being set.

I also tested it.

I inserted this into my page
$numfive = 5;

I inserted this into my include

$numfour = 4;
$done = $numfour + $numfive;

when the page is called it returns "4"

any ideas?