Php read directory file

Question

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.

Test your code with a file named "this_is_text_not_png.txt"

horatio
– horatio

2011-07-08 18:23:50 +00:00
Commented Jul 8, 2011 at 18:23 — horatio
– horatio, Commented Jul 8, 2011 at 18:23

Mark · Accepted Answer · 2012-06-24 21:13:15Z

6

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.'">';
}

edited Jun 24, 2012 at 21:13

answered Jul 8, 2011 at 18:07

Mark

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2 Comments

kwek-kwek Over a year ago

Thank you. I am try to make it load the list randomly by putting shuffle() but I have no luck I am trying to learn PHP as I go along so I do appreciate it.

Alexey Vesnin Over a year ago

If you have an error Directory::read() method not found - try pth_read() instead of read() - undocumented "feature"

rid · Accepted Answer · 2011-07-08 17:40:57Z

2

You'll have a space at the end of the $imglist string, which explode() will turn into an empty element. Trim the string:

$imglist = explode(" ", trim($imglist));

Or better yet, just add them to the $imglist array in the first place, instead of making a string and exploding it:

$imglist = array();
/* ... */
$imglist[] = $file;

answered Jul 8, 2011 at 17:40

rid

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1 Comment

rid Over a year ago

You can use shuffle() to randomize the contents of an array.

Marc B · Accepted Answer · 2011-07-08 17:47:45Z

2

ereg() is deprecated. You'd probably be better off with:

chdir($img_folder);
$imgs = glob('*.jpg *.gif *.png');
foreach ($imgs as $img) {
    echo "<img src=\"{$img_folder}/{$img}\">";
}

glob() does wildcard matching pretty much the same way as most Unix shells do.

answered Jul 8, 2011 at 17:47

Marc B

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1 Comment

Álvaro González Over a year ago

Furthermore, eregi was being used to match substrings anywhere on the file name.

Roshni · Accepted Answer · 2014-06-26 09:09:25Z

0

Use [glob()][1] function
<?php
    define('IMAGEPATH', 'path to my image/'.$imglist.'/');
    foreach(glob(IMAGEPATH.'*.jpg') as $filename){
        echo '<img src="'.$filename.'" alt="'.$album.'" />';

    ?>

  [1]: http://www.php.net/manual/en/function.glob.php

answered Jun 26, 2014 at 9:09

Roshni

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