[PHP] Display random images (from a directory)

Hitch · 03-11-2010, 03:49 PM

There are many scripts and tutorials out there that let you display random images.
Most of them require you to specify which images to show… which can be a pain writing out what images you want to display, especially if there is a lot of them.

Which is why i wrote this PHP function, to make it simpler.
Instead of listing the files, just specify which directory contain the images you want displaying.

This script will “scan” your specified directory, and display the files randomly. (Given they have a image extension of .jpg, .jpeg, .gif, .png, .ico)

The Code

PHP Code:

  <?php

// Written by Cueburst (http://cueburst.com)

error_reporting(E_ALL & ~E_NOTICE);

// Random Image function
function random_image($directory)
{
    $leading = substr($directory, 0, 1);
    $trailing = substr($directory, -1, 1);
    
    if($leading == '/')
    {
        $directory = substr($directory, 1);
    }
    if($trailing != '/')
    {
        $directory = $directory . '/';
    }
    
    if(empty($directory) or !is_dir($directory))
    {
        die('Directory: ' . $directory . ' not found.');
    }
    
    $files = scandir($directory, 1);
    
    $make_array = array();
    
    foreach($files AS $id => $file)
    {
        $info = pathinfo($dir . $file);
    
        $image_extensions = array('jpg', 'jpeg', 'gif', 'png', 'ico');
        if(!in_array($info['extension'], $image_extensions))
        {
            unset($file);
        }
        else
        {
????: NamePros.com http://www.namepros.com/code/644409-php-display-random-images-from-directory.html
            $file = str_replace(' ', '%20', $file);
            $temp = array($id => $file);
            array_push($make_array, $temp);
        }
    }
    
    if(sizeof($make_array) == 0)
    {
        die('No images in ' . $directory . ' Directory');
    }
    
    $total = count($make_array) - 1;

    $random_image = rand(0, $total);
    return $directory . $make_array[$random_image][$random_image];

}

?>

To use the function, it's quite simple, type of your HTML Img line, and add our function in the source, for example, lets say our images are located in a folder called cars...

PHP Code:

  <?php
????: NamePros.com http://www.namepros.com/showthread.php?t=644409

require 'random_image.php';
echo "<img src=" . random_image('cars') . " />";

?>

Final Notes

When entering the directory name, there is no-need for training or leading slashes, the function will check and add them in where applicable.
Accessing parent directories is also allowed, example, ../../cars/
Image's may have spaces in the filenames, they will be replaced with %20.

Enjoy!

Obulus · 04-06-2010, 04:20 AM

Thanks for this!

Nodws · 04-09-2010, 12:08 AM

lol, just name images as numbers. 1.jpg,2.jpg, etc. will save tons on CPU resources

03-11-2010, 03:49 PM	THREAD STARTER #1 (permalink)
Hitch Senior Member Join Date: Aug 2005 Location: Uk, South Yorkshire Posts: 1,228 Follow @adrianc333	[PHP] Display random images (from a directory) There are many scripts and tutorials out there that let you display random images. Most of them require you to specify which images to show… which can be a pain writing out what images you want to display, especially if there is a lot of them. Which is why i wrote this PHP function, to make it simpler. Instead of listing the files, just specify which directory contain the images you want displaying. This script will “scan” your specified directory, and display the files randomly. (Given they have a image extension of .jpg, .jpeg, .gif, .png, .ico) The Code PHP Code: <?php // Written by Cueburst (http://cueburst.com) error_reporting(E_ALL & ~E_NOTICE); // Random Image function function random_image($directory) { $leading = substr($directory, 0, 1); $trailing = substr($directory, -1, 1); if($leading == '/') { $directory = substr($directory, 1); } if($trailing != '/') { $directory = $directory . '/'; } if(empty($directory) or !is_dir($directory)) { die('Directory: ' . $directory . ' not found.'); } $files = scandir($directory, 1); $make_array = array(); foreach($files AS $id => $file) { $info = pathinfo($dir . $file); $image_extensions = array('jpg', 'jpeg', 'gif', 'png', 'ico'); if(!in_array($info['extension'], $image_extensions)) { unset($file); } else { ????: NamePros.com http://www.namepros.com/code/644409-php-display-random-images-from-directory.html $file = str_replace(' ', '%20', $file); $temp = array($id => $file); array_push($make_array, $temp); } } if(sizeof($make_array) == 0) { die('No images in ' . $directory . ' Directory'); } $total = count($make_array) - 1; $random_image = rand(0, $total); return $directory . $make_array[$random_image][$random_image]; } ?> To use the function, it's quite simple, type of your HTML Img line, and add our function in the source, for example, lets say our images are located in a folder called cars... PHP Code: `<?php ????: NamePros.com http://www.namepros.com/showthread.php?t=644409 require 'random_image.php'; echo "<img src=" . random_image('cars') . " />"; ?>` Final Notes When entering the directory name, there is no-need for training or leading slashes, the function will check and add them in where applicable. Accessing parent directories is also allowed, example, ../../cars/ Image's may have spaces in the filenames, they will be replaced with %20. Enjoy!

04-06-2010, 04:20 AM	#2 (permalink)
Obulus NamePros Regular Join Date: Aug 2009 Posts: 313	Thanks for this! __________________ Available for hire.

04-09-2010, 12:08 AM	#3 (permalink)
Nodws Senior Member Join Date: Jun 2008 Location: nodws.com Posts: 1,666 Follow @nodws	lol, just name images as numbers. 1.jpg,2.jpg, etc. will save tons on CPU resources __________________ SnackNames.com \| SnackDesign.com \| buckstabu.com \| adoolt.com \| yoAmoNY.com Developer For Hire! PHP/MySQL/Wordpress/Etc Graphic design, Corporate ID, Dev Ideas

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