How to get random image from directory using PHP

Question

I have one directory called images/tips.

Now in that directory I have many images which can change.

I want the PHP script to read the directory, to find the images, and out of those images found to pick a random image.

Any idea on how to do this?

alex · Accepted Answer · 2012-01-13 14:05:58Z

63

$imagesDir = 'images/tips/';

$images = glob($imagesDir . '*.{jpg,jpeg,png,gif}', GLOB_BRACE);

$randomImage = $images[array_rand($images)]; // See comments

You can send a 2nd argument to array_rand() to get more than 1.

edited Jan 13, 2012 at 14:05

answered Nov 19, 2009 at 6:36

alex

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2 Comments

JAL Over a year ago

Actually, you'd have to do $randomImage=$images[array_rand($images)]; From the manual: "array_rand() returns the key for a random entry."

John Over a year ago

thank buddy , that worked perfectly This webiste is great for getting help. Is there any option that when some comment is posted or answers is given i get email because i had three answers and i didn't get email , i ticked the notify box but still could not get email

Ignas R · Accepted Answer · 2009-11-19 06:28:12Z

4

$images = glob('images/tips/*');
return $images[rand(0, count($images) - 1)];

However, this doesn't ensure that the same image isn't picked twice consecutively.

answered Nov 19, 2009 at 6:28

Ignas R

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2 Comments

wallyk Over a year ago

Adding that insurance would detract from the randomness. It also might introduce bugs.

keithjgrant Over a year ago

I had no clue that glob() function existed. Nice.

Taryn · Accepted Answer · 2012-02-17 01:50:31Z

1

<?php
   foreach (glob("gallery/*") as $filename) {
         echo '<li><a href="'.$filename.'" title=""><img src="'.$filename.'" alt="" /></a> </li>';      
       }
?>

Look at this code, use it definitely if useful for you. It loads all files from folder and prints them in above format. I made this code to use with lightbox.

edited Feb 17, 2012 at 1:50

Taryn

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answered Feb 16, 2012 at 17:02

Makarand Mane

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Comments

Leistungsabfall · Accepted Answer · 2014-10-03 11:08:36Z

1

function get_rand_img($dir)
{
    $arr = array();
    $list = scandir($dir);
    foreach($list as $file)
    {
        if(!isset($img))
        {
            $img = '';
        }
        if(is_file($dir . '/' . $file))
        {
            $ext = end(explode('.', $file));
            if($ext == 'gif' || $ext == 'jpeg' || $ext == 'jpg' || $ext == 'png' || $ext == 'GIF' || $ext == 'JPEG' || $ext == 'JPG' || $ext == 'PNG')
            {
                array_push($arr, $file);
                $img = $file;
            }
        }
    }
    if($img != '')
    {
        $img = array_rand($arr);
        $img = $arr[$img];
    }
    $img = str_replace("'", "\'", $img);
    $img = str_replace(" ", "%20", $img);
    return $img;
}


echo get_rand_img('images');

replace 'images' with your folder.

edited Oct 3, 2014 at 11:08

Leistungsabfall

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answered Oct 3, 2014 at 10:15

aditya7822

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Comments

Sajjad Hossain · Accepted Answer · 2016-01-11 17:49:37Z

1

Agreed with alexa. Use simple function.

function RandImg($dir)
{
$images = glob($dir . '*.{jpg,jpeg,png,gif}', GLOB_BRACE);

$randomImage = $images[array_rand($images)];
return $randomImage;
}

$the_image = RandImg('images/tips/');
echo $the_image;

answered Jan 11, 2016 at 17:49

Sajjad Hossain

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Comments

stalin wesley · Accepted Answer · 2018-04-20 06:01:32Z

1

$folder = "images";
$results_img_arr = array();
if (is_dir($folder))
{
        if ($handle = opendir($folder))
        {
                while(($file = readdir($handle)) !== FALSE)
                {
                    if(!in_array($file,array(".","..")))
                        $results_img_arr[] = $folder."/".$file;
               }
         closedir($handle);
        }
}
$ran_img_key  = array_rand($results_img_arr);

$img_path = $results_img_arr[$ran_img_key];

answered Apr 20, 2018 at 6:01

stalin wesley

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Comments

keithjgrant · Accepted Answer · 2009-11-19 06:28:18Z

0

You can use opendir() to read in the filenames from that directory, storing each filename in an array. Then use rand() with a min and max corresponding to your array keys to select an item from the array.

answered Nov 19, 2009 at 6:28

keithjgrant

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Comments

alex · Accepted Answer · 2014-03-11 09:55:52Z

0

Simpler:

$directory = "medias/photos/";
$img = glob($directory . "*.jpg");
shuffle($img);

edited Mar 11, 2014 at 9:55

alex

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answered Jun 1, 2013 at 7:57

Paolo Euvrard

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1 Comment

alex Over a year ago

Problem is, this would only find jpg images, and it does a lot of unnecessary work by shuffling the whole array (may or may not be a problem, depending on how many images you have in the folder).

Alessandro · Accepted Answer · 2020-01-04 15:43:01Z

I wrote a simple php script for my personal use. Now I want share it with stackoverflow's community. Usage is simple: create a folder "php" into root of your Web Server and put inside this file php rotate.php... now create two folders into your root called "pic" and "xmas"... you can adjust the folder names by editing the var $my_folder_holiday and $my_folder_default...

<?php
  ##########################################################
  # Simple Script Random Images Rotator • 1.4 • 04.01.2020 #
  # Alessandro Marinuzzi [alecos] • https://www.alecos.it/ #
  ##########################################################
  function rotate($folder) {
    if ((file_exists($_SERVER['DOCUMENT_ROOT'] . "/$folder")) && (is_dir($_SERVER['DOCUMENT_ROOT'] . "/$folder"))) {
      $list = scandir($_SERVER['DOCUMENT_ROOT'] . "/$folder");
      $fileList = array();
      $img = '';
      foreach ($list as $file) {
        if ((file_exists($_SERVER['DOCUMENT_ROOT']  . "/$folder/$file")) && (is_file($_SERVER['DOCUMENT_ROOT']  . "/$folder/$file"))) {
          $ext = strtolower(pathinfo($file, PATHINFO_EXTENSION));
          if ($ext == 'gif' || $ext == 'jpeg' || $ext == 'jpg' || $ext == 'png') {
            $fileList[] = $file;
          }
        }
      }
      if (count($fileList) > 0) {
        $imageNumber = time() % count($fileList);
        $img = $folder . '/' . $fileList[$imageNumber];
      }
      return $img;
    } else {
      mkdir($_SERVER['DOCUMENT_ROOT'] . "/$folder", 0755, true);
    }
  }
  $my_gallery_month = date('m');
  $my_folder_default = 'pic';
  $my_folder_holiday = 'xmas';
  if ($my_gallery_month == 12) {
    $my_gallery = rotate($my_folder_holiday);
  } else {
    $my_gallery = rotate($my_folder_default);
  }
?>

This script was tested under PHP 7.0/7.1/7.2/7.3 and PHP 7.4 and works fine. Usage (for example in root you may have a folder "pic" and "xmas" containing your images):

<a href="/<?php include("php/rotate.php"); echo $my_gallery; ?>"><img src="/<?php echo $my_gallery; ?>" alt="Random Gallery" width="90" height="67"></a>

Other usage using FancyBox library:

<a href="/<?php include("php/rotate.php"); echo $my_gallery; ?>" data-fancybox><img src="/<?php echo $my_gallery; ?>" alt="Random Gallery" width="90" height="67"></a>

Hope this Helps.

Dharman · Accepted Answer · 2023-09-19 13:19:26Z

-1

Load folder with images:

$folder = opendir(images/tips/);

Build table out of files/images from directory:

$i = 0;
while(false !=($file = readdir($folder))){
if($file != "." && $file != ".."){
    $images[$i]= $file;
    $i++;
    }
}

Pick random:

$random_img=rand(0,count($images)-1);

Show on page:

echo '<img src="images/tips'.$images[$random_img].'" alt="" />';

edited Sep 19, 2023 at 13:19

Dharman♦

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answered Nov 19, 2009 at 6:30

Czlowiekwidmo

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1 Comment

mpen Over a year ago

Two issues. Don't you need another slash after tips? And why not just do $images[] = $file and forget the counter?

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