Getting file name of image

@michael879Mar 04.2007 — #what are you using the display the slideshow? Whatever it is, you need the filename to be able to display the picture. That means your script knows the filename. Just output it..

@prophitauthorMar 04.2007 — #This is what im using

<i>
 </i>&lt;?php
 $dh = new DirectoryIterator( "pics" );
 
 $files = array();
 foreach( $dh as $file)
 {
 if ( preg_match("/[.]jpg$/", $file)) $files []= "$file";
 }
 ?&gt;
 
 
 &lt;style&gt;
 body		{ background: Black;}
 #thumbnails { height: 140px; width: 100%; overflow: auto;}
 #pic		{ text-align: center; height: 400px; padding: 20px;}
 &lt;/style&gt;
 &lt;script&gt;
 var image_list = [
 &lt;?php $first = true; foreach( $files as $image) { ?&gt;
 &lt;?php echo ($first ? "" : ", "); ?&gt; "&lt;?php echo ($image); ?&gt;"
 &lt;?php $first = false;}?&gt;
 ];
 
 var curimage = 0;
 
 function switchimg( ind )
 {
 var image = image_list[ind];
 var obj = document.getElementById( "selimg" );
 obj.src = "scale.php?image="+image+"&amp;y=400";
 curimage = ind;
 }
 
 function nextimage()
 {
 curimage++;
 if ( curimage &gt;= image_list.length ) curimage = 0;
 switchimg( curimage );
 }
 
 
 window.setInterval( "nextimage()", 10000 );
 &lt;/script&gt;
 
 &lt;table width="100%"&gt;
 &lt;tr&gt;
 &lt;?php $ind = 0; foreach ($files as $image) { ?&gt;
 &lt;td width="160" nowrap align="center"&gt;
 &lt;a href="javascript:switchimg( &lt;?php echo ($ind); ?&gt; )"&gt;
 &lt;img height="100" src="scale.php?image=&lt;?php echo ($image); ?&gt;&amp;y=100" border="0" /&gt;
 &lt;/a&gt;
 &lt;/td&gt;
 &lt;?php $ind++; } ?&gt;
 &lt;/tr&gt;
 &lt;/table&gt;
 &lt;table&gt;
 &lt;tr&gt;
 &lt;td width="830"&gt;
 &lt;div id="pic"&gt;
 &lt;img id="selimg" height="400" src="scale.php?image=&lt;?php echo ($files[0]);?&gt;&amp;y=400" /&gt;
 &lt;br&gt;
 &lt;?php
 
 echo '&lt;font color="ffffff"&gt;Filename&lt;/font&gt;';
 ?&gt;
 &lt;br&gt;
 &lt;/div&gt;
 &lt;/td&gt;
 &lt;/tr&gt;
 &lt;/table&gt;

the script uses javascript and php to change the image. I'm just not sure how to pull out the variables to echo or print the filename of the displaying image.

@michael879Mar 04.2007 — #scale.php?image=<?php echo ($image); ?>&y=100

that is the images url..

Getting file name of image

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