Probelm displaying files from directory

PHP
1

Hello folks,

Please view the following page: http://www.roy.lu/futuraplay/

I am using this code:


$dir="./download"; // Directory where files are stored

if ($dir_list = opendir($dir))
{
while(($filename = readdir($dir_list)) !== false)
{
?>
<p><a href="<?php echo $filename; ?>"><?php echo $filename;
?></a></p>
<?php
}
closedir($dir_list);
}

1. Why does this display a . and a … in addition to the files in the directory? - I don’t want this
2. Why don’t the links include the real file path? Hence the pictures won’t open…

Thanks in advance for your help.

Roy

2

Because you didn’t ask it to differentiate between files and directories. Each folder in linux contains a . and … for current folder and parent folder. You can use is_file($dir . ‘/’ . $filename) to get rid of them.

Because $filename is just the file name. You need to use $dir . ‘/’ . $filename to get the full path.

3

Ok, I understand what you are saying, but I can’t figure out where/ how i need to change my code.
A little help please?

4

Certainly,

<?php
$dir="./download"; // Directory where files are stored 

if ($dir_list = opendir($dir)) 
{ 
   while(($filename = readdir($dir_list)) !== false) 
   {
      $pathToFile = $dir . '/' . $filename;
      if (is_file($pathToFile)) // added this and changed the link to use $pathToFile
      { 
?> 
   <p><a href="<?php echo $pathToFile; ?>"><?php echo $filename; ?></a></p> 
<?php 
      }
   } 
   closedir($dir_list); 
}
?>
5

Thanks a lot, I did manage to get the link working properly, but I’m not familiar enough with PHP yet to do this kind of magic…
I was trying something along the lines of:

if (($dir_list = opendir($dir)) && is_file($dir . ‘/’ . $filename))
{ …

but this wasn’t as logical to PHP as it seemed to me :wink: