PHP: display most recent images from directory?
I have a camera server FTPing images to a webserver. Can anyone suggest the PHP snippet I'd need that would look th开发者_JS百科rough the server's public root directory (/public_html) and display the four most recent images?
I can tell the camera server to name uploaded images by date/time, however needed [eg. image-021020102355.jpg
for an image created 2nd oct 2010 at 11:55pm]
thanks!
I've put together something that can help you. This piece of code displays the last recent images in the root directory of your server.
<?php
$images = glob('*.{gif,png,jpg,jpeg}', GLOB_BRACE); //formats to look for
$num_of_files = 4; //number of images to display
foreach($images as $image)
{
$num_of_files--;
if($num_of_files > -1) //this made me laugh when I wrote it
echo "<b>".$image."</b><br>Created on ".date('D, d M y H:i:s', filemtime($image)) ."<br><img src="."'".$image."'"."><br><br>" ; //display images
else
break;
}
?>
This should do it:
<?php
foreach (glob('*.jpg') as $f) {
# store the image name with the last modification time and imagename as a key
$list[filemtime($f) . '-' . $f] = $f;
}
$keys = array_keys($list);
sort($keys); # sort is oldest to newest,
echo $list[array_pop($keys)]; # Newest
echo $list[array_pop($keys)]; # 2nd newest
If you can make the filenames YYYYMMDDHHMM.jpg sort() can put them in the right order and this would work:
<?php
foreach (glob('*.jpg') as $f) {
# store the image name
$list[] = $f;
}
sort($list); # sort is oldest to newest,
echo array_pop($list); # Newest
echo array_pop($list); # 2nd newest
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