php script to show 4 random photos from a folder is showing the same photo 4 times

Question

i'm trying to build a little module for a site that pulls 4 random photos from a folder on the server and presents them in a div. what's happening is the same 4 photos are being called. how do i get it to call 4 different photos?

heres the random.php file:

<?php
$folder = '.';

$extList = array();
$extList['jpg'] = 'image/jpeg';
$extList['jpeg'] = 'image/jpeg';

$img = null;

if (substr($folder,-1) != '/') {
$folder = $folder.'/';
}

if (isset($_GET['img'])) {
$imageInfo = pathinfo($_GET['img']);
if (
    isset( $extList[ strtolower( $imageInfo['extension'] ) ] ) &&
    file_exists( $folder.$imageInfo['basename'] )
) {
    $img = $folder.$imageInfo['basename'];
}
} else {
$fileList = array();
$handle = opendir($folder);
while ( false !== ( $file = readdir($handle) ) ) {
    $file_info = pathinfo($file);
    if (
        isset( $extList[ strtolower( $file_info['extension'] ) ] )
    ) {
        $fileList[] = $file;
    }
}
closedir($handle);

if (count($fileList) > 0) {
    $imageNumber = time() % count($fileList);
    $img = $folder.$fileList[$imageNumber];
}
}

if ($img!=null) {
$imageInfo = pathinfo($img);
$contentType = 'Content-type: '.$extList[ $imageInfo['extension'] ];
header ($contentType);
readfile($img);
} else {
if ( function_exists('imagecreate') ) {
    header ("Content-type: image/png");
    $im = @imagecreate (100, 100)
        or die ("Can't initialize image stream");
    $background_color = imagecolorallocate ($im, 255, 255, 255);
    $text_color = imagecolorallocate ($im, 0,0,0);
    imagestring ($im, 2, 5, 5,  "IMAGE ERROR", $text_color);
    imagepng ($im);
    imagedestroy($im);
}
} 

?>

here's how random.php is being called:

<?php

/* DATA */
$data = array(
array('COUNT', '88', 'Here are a few of them', '1, 2, 3, 4'),
);
?>

<div>
<table class="reop"  border='0' width='100%'  cellpadding='0' cellspacing='10'>

<?php
$count = 0;
foreach($data as $row) {
    $class = ($count % 2 == 1 ? " class='alt'" : '');
    echo "<tr$class>";
    for($j = 0; $j < count($row); $j++) {
        if ($j!=3) {
            echo "<td class='cell_$j'>$row[$j]</td>";
        } else {


 //           $avatar = ''; 
            $array = preg_split('/,/', $row[$j], -1, PREG_SPLIT_NO_EMPTY);
            foreach ($array as $val) {
            $avatar .= '<img src="/staffpics/random.php"> ';
            }   
            echo "<td class='cell_$j'>$avatar</td>";

        }   

    }   
    echo '</tr>';
//    $count++;
}   

?>  

</table>

I don't have an answer, but I do know why the 4 pictures come out the same: your browser caches all images. Once it has loaded an image from the server, it sees no need to load the identical image (i.e. with the same filename) again. So maybe give them 4 different names? — Mr Lister, Dec 20 '11 at 19:38
@MrLister - hmm, all the images in /staffpics/ are named [1.jpg, 2.jpg. 3.jpg, ...] is this what you mean? — tronjavolta, Dec 20 '11 at 19:41
No, not the jpegs themselves, the filenames that the browser sees them as. "/staffpics/random.php", that's the name that should vary. — Mr Lister, Dec 20 '11 at 20:35
After a lot of searching, this answer was the best for me. Emailing bday cards to my talents on their day. Have folder with 100 pictures that I can add/delete without having to rename. Works perfectly. http://stackoverflow.com/a/10262008/2150167 Look for Razon's answer. — cliffordem, Apr 29 '13 at 00:41

Kaii · Accepted Answer · 2011-12-21T00:35:48.163

when your page is displayed, most browsers will immediately request the four pictures.. nearly all at the same time. your code for "random" picking a picture is based on time() which returns a count of seconds since the begin of unixtime. executed 4 times within the same second it will also yield the same result 4 times.

your code:

$imageNumber = time() % count($fileList);
$img = $folder.$fileList[$imageNumber];

should be rewritten as:

// filelist starts with "0", count with "1" .. so we are off-by-one and first picture would never show without -1
$imageNumber = rand() % (count($fileList) - 1);
$img = $folder.$fileList[$imageNumber];

that way you will get "random" pictures.

Also, what Mister Lister commented is totally valid - your browser may also cache the picture and thus only request a picutre once and just display that one again and again using its cache.

to avoid this, replace any <img src="random.php"> with <img src="random.php?rand=<?= rand(); ?>">

But note:

your server may eventually return the same random number multiple times so eventually (rarely) you will end up with the same picture twice on the same page.

to fix that, you must completely rewrite your code.

the script rendering the page should decide which pictures to choose and ensure that no picture is displayed twice instead of just including a "give random picture" script, which is independent of the page rendered.

much easier, you could try this:

// find all images - case SenSItivE
$all_images = glob("/path/to/images/*.{jpeg|jpg|png|gif}", GLOB_BRACE);
// bring array in random order
shuffle($all_images);
// pick four random images - you may also use a for or foreach loop to iterate the array
list ($img1, $img2, $img3, $img4) = $all_images; 
// write code to display four images here

Oops, I didn't even see that the OP had no rand() in his code! — Mr Lister, Dec 21 '11 at 13:20

score 1 · Answer 2 · answered Dec 20 '11 at 20:50

1

Here is another way of fetching 4 random photos:

//randomize order
shuffle($fileList);
//get first 4 values of array
$randomPhotos = array_slice($fileList, 0, 4);

Please note, shuffle would change value order in your original array.

answered Dec 20 '11 at 20:50

Nazariy

6,028
5
37
61

score 0 · Answer 3 · answered Dec 20 '11 at 19:37

0

Call srand to seed your random number generator.

answered Dec 20 '11 at 19:37

user229044

232,980
40
330
338

php script to show 4 random photos from a folder is showing the same photo 4 times

3 Answers3