问题描述
我有一个摄像头服务器将图像通过 FTP 传输到网络服务器.任何人都可以建议我需要的 PHP 代码片段来查看服务器的公共根目录 (/public_html) 并显示最近的四个图像吗?
I have a camera server FTPing images to a webserver. Can anyone suggest the PHP snippet I'd need that would look through the server's public root directory (/public_html) and display the four most recent images?
我可以告诉相机服务器按日期/时间命名上传的图像，但需要[例如.image-021020102355.jpg 用于 2010 年 10 月 2 日晚上 11:55 创建的图像]
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]
谢谢！
推荐答案
应该这样做:
<?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
如果您可以使文件名 YYYYMMDDHHMM.jpg sort() 可以将它们按正确的顺序排列，这将起作用:
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
这篇关于PHP:显示目录中的最新图像?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！