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1. jderosa3
   Member
   Posted 3 years ago #

   I would like to use a simple PHP script to show all of the images in a directory

   <?php
   	$handle = opendir ('http://www.xxx.com/flash/portfolio/branding/images/');
       while (false !== ($file = readdir($handle))) {
           echo '<img src="'.$file.'"/><br />'.$file.'<br />';
       }
   ?>

   This throws back a couple of errors:

   Warning opendir(http://www.xxx.com/flash/portfolio/branding/images/)
   [function.opendir]: failed to open dir: not implimented in.... "location of exec-php plugin"

   Any ideas how to get a simple script like this to work?

2. Otto
   Tech Ninja
   Posted 3 years ago #

   Well, writing it so that it's not broken code would be a good start.

   The PHP command opendir is trying to open a directory up for reading. However, instead of giving it a directory, you gave it a URL. I wouldn't expect that to actually work because a URL is not a directory.

3. jderosa3
   Member
   Posted 3 years ago #

   I have tried changing

   this:
   

   <?php
   	$handle = opendir ('http://www.xxx.com/flash/portfolio/branding/images/');
       while (false !== ($file = readdir($handle))) {
           echo '<img src="'.$file.'"/>'.$file.'';
       }
   ?>

   to this:
   

   <?php
   	$handle = opendir ('../flash/portfolio/branding/images/');
       while (false !== ($file = readdir($handle))) {
           echo '<img src="'.$file.'"/>'.$file.'';
       }
   ?>

   No luck.. is there a javascript snippet that does what I want?

   These above still cause errors with the exec-PHP plugin... not sure what going on...

4. jderosa3
   Member
   Posted 3 years ago #

   Ok... I finally have it reading file onto the page, but... they come up with broken icons as if the graphics are not there... here is my new code:

   <?php
   $dir = "./flash/portfolio/branding/images/";
   //open dir
   if ($opendir = opendir($dir))
   {
      //read dir
   while (($file = readdir($opendir)) !== FALSE)
   {
   if ($file!="."&amp;&amp;$file!="..")
   echo $file."<img src='$dir/$file'><br />";
      }
   }
   ?>

   Any ideas?

5. jderosa3
   Member
   Posted 3 years ago #

   got it to work! Incase anyone needs to know how to do this:
   

   <?php
   $dir = "./flash/portfolio/branding/images/";
   $jpgs = glob($dir . "*.jpg");
   
   foreach ($jpgs as $jpg) {
       $jpg = substr($jpg, 1);
       echo "<img src='http://www.mydomain.com{$jpg}' /><br />";
   }
   ?>

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