PHP脚本显示文件夹中的图像

Question

嗨，我有安全摄像头，通过ftp上传到我的服务器，我想显示最后的图像作为幻灯片，但我不能管理它的工作。我有这个代码

$base_path = 'wp-content/uploads/camer/10.121.0.202';
$latest_date_folder = scandir($base_path, SCANDIR_SORT_DESCENDING);
$latest_folder = scandir($base_path  . "/" . $latest_date_folder[0], SCANDIR_SORT_DESCENDING);
$directory = '../" . $base_path  . "/" . $latest_date_folder[0] . "/" . $latest_folder[0] . ';     
try {       

echo '<div id="myslides">'; 
foreach ( new DirectoryIterator($directory) as $item ) {            
if ($item->isFile()) {
$path = $directory . '/' . $item;   
echo '<img src="' . $path . '"/>';  
}
}   
echo '</div>';
}   
catch(Exception $e) {
echo 'No images found for this slideshow.<br />';   
}
?> 

我仍然得到"没有找到此幻灯片的图像"。但是当我尝试这段代码

$latest1_date_folder = scandir($base1_path, SCANDIR_SORT_DESCENDING);
$latest1_folder = scandir($base1_path  . "/" . $latest1_date_folder[0], SCANDIR_SORT_DESCENDING);
$latest1_file = scandir($base1_path  . "/" . $latest1_date_folder[0] . "/" . $latest1_folder[0] , SCANDIR_SORT_DESCENDING);
echo "<img src='../" . $base1_path  . "/" . $latest1_date_folder[0] . "/" . $latest1_folder[0] . "/" . $latest1_file[0] . "' />";

正常显示最后一张图像。我做错了什么?非常感谢。

我正在使用wordpress插件phpcode片段

Answer 1

您的目录变量错误!你应该这样写

$directory = '../'. $base_path  . "/" . $latest_date_folder[0] . "/" . $latest_folder[0];