Pull all images from a specified directory and then display them

Question

How to display the image from a specified directory? like i want to display all the png images from a directory, in my case my directory is media/images/iconized.

I tried to look around but seems none of them fits what i really needed.

But here's my try.

$dirname = "media/images/iconized/"; $images = scandir($dirname); $ignore = Array(".", ".."); foreach($images as $curimg){     if(!in_array($curimg, $ignore)) {         echo "<img src='media/images/iconized/$curimg' /><br>\n";     } } 

hope someone here could help. Im open in any ideas, recommendation and suggestion, thank you.

Answer 1

You can also use glob for this:

$dirname = "media/images/iconized/"; $images = glob($dirname."*.png");  foreach($images as $image) {     echo '<img src="'.$image.'" /><br />'; } 

Answer 2

You can display all image from a folder using simple php script. Suppose folder name “images” and put some image in this folder and then use any text editor and paste this code and run this script. This is php code

    <?php      $files = glob("images/*.*");      for ($i=0; $i<count($files); $i++)       {         $image = $files[$i];         $supported_file = array(                 'gif',                 'jpg',                 'jpeg',                 'png'          );           $ext = strtolower(pathinfo($image, PATHINFO_EXTENSION));          if (in_array($ext, $supported_file)) {             echo basename($image)."<br />"; // show only image name if you want to show full path then use this code // echo $image."<br />";              echo '<img src="'.$image .'" alt="Random image" />'."<br /><br />";             } else {                 continue;             }           }        ?> 

if you do not check image type then use this code

<?php $files = glob("images/*.*"); for ($i = 0; $i < count($files); $i++) {     $image = $files[$i];     echo basename($image) . "<br />"; // show only image name if you want to show full path then use this code // echo $image."<br />";     echo '<img src="' . $image . '" alt="Random image" />' . "<br /><br />";  } ?>