如何存储与数据库链接的文件的最佳方式,其中没有太多文件存储在同一目录中(减慢FTP浏览速度等)
我自己的建议:
3000000/0/70000/9000/100/30/3079138.full.pdf
3000000/0/70000/9000/100/30/3079138.thumb.png
3000000/0/70000/8000/900/20/3078928.full.png
3000000/0/70000/8000/900/20/3078928.thumb.pdf
3000000/0/70000/8000/900/20/3078929.full.pdf
3000000/0/70000/8000/900/20/3078929.thumb.png
答案 0 :(得分:1)
使用此解决方案,每个目录中最多总共有10个文件(id)
对于从1到1999的文件(id),根目录中的目录将如下所示:
require_once 'File_structure.php';
for($i=1; $i<2000; $i++){
$File = new File_structure();
$File->id = $i;
$File->prepend_path = 'image/';
$path = $File->create();
file_put_contents($path.$i, 'sd');
}
require_once 'File_structure.php';
$id = 1254;
$File = new File_structure();
$File->id = $id;
$File->prepend_path = 'image/';
$path = $File->get();
if(is_file($path.$id)){
echo 'ok';
}
class File_structure {
public $id;
public $prepend_path = '';
private $path = array();
function get(){
$this->path();
return $this->prepend_path.(substr($this->prepend_path, -1) == '/' ? '':'/').implode('/', $this->path).'/';
}
function create(){
$this->path();
$path = substr($this->prepend_path, -1) == '/' ? substr($this->prepend_path, 0, -1):$this->prepend_path;
foreach($this->path as $dir){
$path .= '/'.$dir;
if(!is_dir($path)){
mkdir($path);
chmod($path, 0777);
}
}
return $path.'/';
}
function path(){
$this->prepare_id();
$length = strlen($this->id);
for($i=0; $i<$length; $i++){
$len = $length - $i;
if($len <= 1){
break;
}
if($path = (int)str_pad($this->id[$i], $len, 0, STR_PAD_RIGHT)){
$this->path[] = $path;
}
}
}
function prepare_id(){
$this->id = (string)$this->id;
}
}
答案 1 :(得分:0)
通常我会根据易于使用的简单模式存储文件:
uploads/<year>/<month>/<day>/full.pdf
uploads/<year>/<month>/<day>/thumb.png
我自然检查文件名冲突并纠正它们