Code: Select all
,Code: Select all
and [syntax="..."] tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read: [url=http://forums.devnetwork.net/viewtopic.php?t=21171]Posting Code in the Forums[/url] to learn how to do it too.[/color]
I am having problems displaying the actual image using directory functions. This is the Code:Code: Select all
$path = "Bobsimages/$event/";
//using the opendir function
$dir_handle = @opendir($path) or die("Unable to open $path");
//running the while loop
while ($file = readdir($dir_handle)) {
if ($file!="." && $file!=".." && $file!="Thumbs.db"){ // <a href='$file'>$file</a><br/>
echo("<a href='#' onclick=showPreview('Bobsimages/$event/$file','1'); return false\"> $file </a>
<img src=\"Bobsimages/$event/$file width=\"95%\" target = \"gol\">");
}
}
closedir($dir_handle);On the left is where the scrollbar is and when you click on the image it displays on a screen on the right. I am not using frames (FYI). This all works fine, When I click on the name of the image it does display the right image to the right of it. But I dont want the names to appear I want the image itself.
Is there a directory function that displays the actual image rather then the name of the image? If so then what? If not then how would I fix this problem?
Can anyone help me with this issue please.
Thanks in advance,
Brian Fonseca
feyd | Please use
Code: Select all
,Code: Select all
and [syntax="..."] tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read: [url=http://forums.devnetwork.net/viewtopic.php?t=21171]Posting Code in the Forums[/url] to learn how to do it too.[/color]