I use this function to get directory size and store total on MySQL with Cronjob
(function(doc) {})(document);
On large directory > 30 GB and hundreds of files, is very slow an load 100% CPU
Exist an equivalent function to get directory size without load CPU 100% and most quickly?
TKS HID
答案 0 :(得分:1)
以这种方式获得尺寸:
chdir ($DirectoryPath);
$bytes = intval(preg_replace('/[\D]+/','',system('du',$result)));
这将获取目录和子目录大小的详细信息。
system('du',$result)
这将删除除数字
以外的所有字符preg_replace('/[\D]+/','',
这会将返回的数字字符串转换为整数值。
intval()