How do I list imags from directory? - php

Question

I'm trying to show all images within a specified directory.
The following code lists all allowed file names:
function getDirectoryList()
{
// create an array to hold directory list
$results = array();
// create a handler for the directory
$handler = opendir($this->currentDIR);
// open directory and walk through the filenames
while ($file = readdir($handler)) {
// Make sure we get allowed images types
if ($this->allowedFileType($file,$this->allowedImageTypes) )
{
$results[] = $file;
}
}
// tidy up: close the handler
closedir($handler);
// done!
return $results;
}
(...)
$images = getDirectoryList();
foreach($images as $img) {
echo "<li>".$img."</li>";
}
How can I get file size and MIME type?
I read that mime_content_typeis deprecated and I should use finfo_file istead. But I've not been very successfull with this.
Do I have to use /usr/share/misc/magic to get file information? Can't I use GD library?
I've looked at many examples, but they are old and don't work that well.
Any help appreciated.

Answer 1

to get the size and mime type of image its simple,
use function : getimagesize
uses like :
list($width, $height, $type, $attr) = getimagesize("img/myimg.jpg");
Returns an array with 7 elements.
Index 0 and 1 contains respectively the width and the height of the image.
Index 2 is one of the IMAGETYPE_XXX constants indicating the type of the image.
using filesize give the size in bytes

Answer 2

To expand on Haim Evgi's post, use getimagesize() to retrieve the dimensions and the image type in an array. Then, use image_type_to_mime_type() on the image type code to retrieve the MIME:
list ($fileWidth, $fileHeight, $fileType) = getimagesize($filename);
$fileMimeType = image_type_to_mime_type($fileType);