Show random image(s) from a directory

Posted on September 21, 2008, under PHP 

Hi,

If you need to show images randomly from a directory then this script can help you.

index.php

Let’s configure some variables: $extensions, $images_folder_path & $url_to_images_folder.

<?php
// extensions to be checked

$extensions = array('jpg','jpeg','gif','png','bmp');

// images folder

$images_folder_path = '/home/site/public_html/images/';

// url to folder

$url_to_folder = 'http://www.yoursitename.com/images/';

Create the array with the URLs to the images:

// Images Array (SRC value)

$images = array();

srand((float) microtime() * 10000000); // IF PHP Version < 4.2.0

// Open directory and read images

if ($handle = opendir($images_folder_path)) {
    while (false !== ($file = readdir($handle))) {
        if ($file != "." && $file != "..") {
            
			// get file extension
			$ext = strtolower(substr(strrchr($file, "."), 1));

			if(in_array($ext, $extensions))
			{
			$images[] = $url_to_folder.$file;
			}

        }
    }
    closedir($handle);
}

Are there any images in the directory? Select randomly 2 distinctive values. If there are no images in the folder, notify the user with a message.

if(!empty($images)) // Do we have something to show?
{
$rand_key = array_rand($images, 1);

$src = $images[$rand_key];

echo "<img src='".$src."' align='absmiddle'>";

/* 
Show second image; Make sure it will not be the same as the first one;
We will remove the element of the first image from the array. This way the script will not reselect it 
*/

unset($images[$rand_key]);

$rand_key = array_rand($images, 1);

$src = $images[$rand_key];

echo "<br /><br /><img src='".$src."' align='absmiddle'>";
}
else
{
echo 'No images were found in <strong>'.$images_folder_path.'</strong>';
}
?>


  • September 21, 2008
  • article by Gabriel C.
  • 6 comments

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6 Replies to "Show random image(s) from a directory"

  1. Tim
    September 21, 2008 at 6:29 AM
    Reply

    This could be done with a lot less code.

  2. Gabriel
    September 21, 2008 at 6:44 AM
    Reply

    A lot less code? Could be, but think how many beginners would actually understand a similar script with less code. The current application is well commented and gives the user an idea of how this script is actually working (step by step).

  3. Chris
    December 21, 2009 at 1:52 PM
    Reply

    Good basic script.

    You can: Make your rand_key array equal to the number of images that you want to show, then create a loop to display each image. This simple change makes this much more versatile.

    Cheers

  4. Roman
    October 29, 2010 at 2:05 AM
    Reply

    Hello, does anybody know how to show 6 random images? Thanks for a help.

  5. Rick de Graaff
    December 23, 2010 at 2:44 PM
    Reply

    Sorry but it can be better! Nice script, made some tweaks.
    It now gets every photo out of the folder.. and shows them

    <?php
    # Credits: Rickdegraaff.nl

    // extensions to be checked

    $extensions = array('jpg','jpeg');

    // images folder

    $images_folder_path = 'images/projects/';

    // url to folder

    $url_to_folder = 'http://'.$_SERVER['SERVER_NAME'].'/porto/images/projects/';

    // Images Array (SRC value)

    $images = array();

    srand((float) microtime() * 10000000); // IF PHP Version < 4.2.0

    // Open directory and read images

    if ($handle = opendir($images_folder_path)) {
    while (false !== ($file = readdir($handle))) {
    if ($file != "." && $file != "..") {

    // get file extension
    $ext = strtolower(substr(strrchr($file, "."), 1));

    if(in_array($ext, $extensions))
    {
    $images[] = $url_to_folder.$file;
    }

    }
    }
    closedir($handle);
    }
    foreach ($images as $imgvalue) {
    print '
    ';
    }
    ?>

    1. Gabriel C.
      January 3, 2011 at 5:46 AM
      Reply

      The part:

      foreach ($images as $imgvalue) {
      print '
      ';
      }
      i

      is incomplete. Make sure you make the code postable first before submitting it inside a comment: http://www.elliotswan.com/postable/

      Thanks for your suggested code!

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