Problem showing 2 ads

Have changed the code to:
<?php
ini_set ("include_path", ini_get ("include_path") . ':../:../../:../../../:../../../../');
include ('../ad_network.php');
echo $ad_network[0],$ad_network[1];
?>

But only one advert showing. What have I done wrong?

 

This is how I have done it from the beginning:

<?php
ini_set ("include_path", ini_get ("include_path") . ':../:../../:../../../:../../../../');
include ('ad_network.php');
echo $ad_network[0];
?> |
<?php
ini_set ("include_path", ini_get ("include_path") . ':../:../../:../../../:../../../../');
include ('ad_network.php');
echo $ad_network[1];
?> |
<?php
ini_set ("include_path", ini_get ("include_path") . ':../:../../:../../../:../../../../');
include ('ad_network.php');
echo $ad_network[2];
?> |
<?php
ini_set ("include_path", ini_get ("include_path") . ':../:../../:../../../:../../../../');
include ('ad_network.php');
echo $ad_network[3];
?> |
<?php
ini_set ("include_path", ini_get ("include_path") . ':../:../../:../../../:../../../../');
include ('ad_network.php');
echo $ad_network[4];
?>

Which puts a pipe in between as well

 

Ok - thanks

 

Not the most beautiful way Foxy did it but it works...

My code to show 5 in PHP:

ini_set ("include_path", ini_get ("include_path") . ':../:../../:../../../:../../../../');
	include ('ad_network.php');
    echo "<p>Some other internet browsing ideas:<br>";
	echo ' '.$ad_network[0].' / '.$ad_network[1].' / '.$ad_network[2].' / '.$ad_network[3].' / '.$ad_network[4].'</p>';

PHP:

With slashes bewteen them.

Should work including just once if you use the latest code.