Visualize a file from a directory

[Daddy]

Hi,

 

I have this code

<?php
 // Must include this
include("geoiploc.php");

function getYJLINKS($default_font_family,$yj_copyrightear,$yj_templatename,$show_tools,$show_fres,$show_rtlc,$validators_off,$totop_off){
	
	$allLinks = array();

	if((function_exists('toolbox_urls') && $show_tools == 1) || $validators_off == 1 || $totop_off == 1){
		echo '<div class="validators">';
		if($validators_off == 1){
			$allLinks[] ='<a href="http://jigsaw.w3.org/css-validator/check/referer?profile=css3" target="_blank" title="CSS Validity">CSS Valid</a>';
			$allLinks[] ='<a href="http://validator.w3.org/check/referer" target="_blank" title="XHTML Validity">XHTML Valid</a>';
		}
		if($totop_off == 1){
			$allLinks[] ='<a class="yjscroll" href="#stylef'.$default_font_family.'">Top</a>';
		}
		if (function_exists('toolbox_urls') && $show_tools == 1):
		global $font_size;
		global $font_direction;
			if ($show_fres == 1):
				$allLinks[] = '<a id="fontSizePlus" class="fs" href="javascript:;" rel="nofollow">+</a>';
				$allLinks[] = '<a id="fontSizeMinus"  class="fs" href="javascript:;" rel="nofollow">-</a>';
				$allLinks[] = '<a id="fontSizeReset"  class="fs" href="javascript:;" rel="nofollow">reset</a>';
			endif;
			if ($show_rtlc == 1):
				$allLinks[] = '<a class="tdir" href="'.$font_direction[1].'" rel="nofollow">RTL</a>';
				$allLinks[] = '<a class="tdir" href="'.$font_direction[2].'" rel="nofollow">LTR</a>';
			endif;
		endif;
		echo implode(' | ',$allLinks);
		echo '</div>';
   }
// loading this from a database
    $ip = $_SERVER["REMOTE_ADDR"];

// listing Country Flags Images
    $dir = 'images/flags';
    $file_display = array('png');
         if (file_exists ($dir) == false) {
    echo 'Directory \'', $dir, '\' not found!';
    }
    else{
         $dir_contents = scandir($dir);
         foreach($dir_contents as $file) {
         $file_type = strtolower(end(explode('.', $file)));
    If($file !== '.' && $file !== '..' && in_array($file_type, $file_display) == true)
    if($file == '..' || $file == '.') {
    continue;
    }
         $code = strtolower(getCountryFromIP($ip, "code"));
    echo '<div class="yjsgcp">Palana.it ® 1994 - '.$yj_copyrightear.' :: Tutti i diritti riservati :: Realizzazione Palana.it &boxv;&boxv; Utente rilevato: IP: '. $ip . ', '.getCountryFromIP($ip, " NamE ").' ('.getCountryFromIP($ip, "code").'-'.getCountryFromIP($ip, "AbBr").') ';
    echo '<img src="', $dir, '/', $code, '.png" /> </div>';
        }
    }
}
?>

The problem, as we see (http://www.palana.it), is that code printed the Italian flag but many times there are images in the directory, that is over 204. See bottom of the page. How can I fix the code to get ONE display?

[Ch0cu3r]

You are getting the ip address line repeated because the code that outputs it is within a foreach loop. It will output the ipaddress for every file within the images/flags/ directory. There is no need for the foreach loop

[Daddy]

Now it works fine: http://www.palana.it

 

Many thanks .