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Home/ Questions/Q 6108895
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T14:20:41+00:00 2026-05-23T14:20:41+00:00

I have a script that goes through a directory that has 3 images $imglist=”;

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I have a script that goes through a directory that has 3 images

$imglist='';
$img_folder = "path to my image";

//use the directory class
$imgs = dir($img_folder);

//read all files from the  directory, checks if are images and ads them to a list 
while ($file = $imgs->read()) {
  if (eregi("gif", $file) || eregi("jpg", $file) || eregi("png", $file))
    $imglist .= "$file ";
} 
closedir($imgs->handle);

//put all images into an array
$imglist = explode(" ", $imglist);

//display image
foreach($imglist as $image) {
  echo '<img src="'.$img_folder.$image.'">';
}

but the problem that I am having is it display a 4th img with no image.. yet I only have 3 image in that folder.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T14:20:42+00:00Added an answer on May 23, 2026 at 2:20 pm

    There is no need to build a string of images and then explode that string into an array of images instead just add the images directly to an array as Radu mentioned.
    Here is the corrected code:

    $imglist = array();
    $img_folder = "path to my image";
    
    //use the directory class
    $imgs = dir($img_folder);
    
    //read all files from the  directory, checks if are images and adds them to a list 
    while ($file = $imgs->read()) {
       if (eregi("gif", $file) || eregi("jpg", $file) || eregi("png", $file)){
          $imglist[] = $file;
       } 
    }
    closedir($imgs->handle);
    
    //display image
    foreach($imglist as $image) {
        echo '<img src="'.$img_folder.$image.'">';
    }
    
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