如何重命名压缩文件名？

Posted 2023-03-07

【中文标题】如何重命名压缩文件名？

【英文标题】：How to rename zipped filename?

【发布时间】：2020-12-22 09:09:43

【问题描述】：

所以我有代码可以从 1 个目录移动到另一个目录文件并压缩该文件。

我需要的： 将压缩文件名重命名为第二个“-”符号。

示例：我得到了压缩文件名“SOMETEXT-de_dust2-20123323.dem.zip”。我需要该文件名只是“SOMETEXT.dem.zip” 所以只需删除所有文本，直到第二个 -"-"

有什么建议吗？ 感谢您帮助我理解代码:)

我的代码：

<?php
//error_reporting(E_ALL);
//set_time_limit(0);
$path = "MIX1/cstrike";
$path2 = "/var/www/html/public/";
$to_dirs = array('/demos/');
$from_dirs = array('/demos/');

$filesizes = array();
//первый проход запоминаем размеры
foreach($from_dirs as $from_dir)
    
    $demos_dir = opendir($path.$from_dir);
    while (false!==($file=readdir($demos_dir)))
        
        if ($file!='.'&&$file!='..'&&strpos($file,'.dem')!==false)
            
            $fsize=filesize($path.$from_dir.$file);
            if ($fsize<50000000)
                
                $filesizes[$file]=$fsize;
                
            else
    //          echo "<br/>bad file:",$file, ",  size = ", $fsize;
                
            
        
    closedir($demos_dir);
    
//echo date("h:i:s");
sleep(3);
clearstatcache ();
//второй проход пермещаем 
$i=0;
foreach($from_dirs as $from_dir)
    
    $to_dir=$from_dirs[$i];
    $demos_dir = opendir($path.$from_dir);
    while (false!==($file=readdir($demos_dir)))
        
        if ($file!='.'&&$file!='..'&&strpos($file,'.dem')!==false)
            
            $fsize=0;
            $fsize=filesize($path.$from_dir.$file);
            if ($fsize<50000000)
                
                if ($fsize==$filesizes[$file])
                    
                    //echo "<br>ѕеремещаем файл ",$file," размер не изменилс¤; было ",$filesizes[$file]," стало, ".$fsize,";";
                    move_demo($file, $from_dir, $to_dir);
                    
                else
                    
                    //echo "<br>","размер изменилс¤ у файла ", $file;
                    
                
            else
                
                //echo "<br/>bad file:",$file, ",  size = ", $fsize;
                
            
        
    $i++;
    closedir($demos_dir);
    

function move_demo($filename, $from_dir, $to_dir)

//echo $filename,"from ",$from_dir," to ",$to_dir,"<br>";
global $path, $path2;
if (file_exists($path2.$to_dir.$filename.".zip"))
    unlink($path2.$to_dir.$filename.".zip");
echo "$path$from_dir$filename\n";
echo "$path2$to_dir$filename\n\n";
$data = file_get_contents($path.$from_dir.$filename);
$gzdata = gzencode($data, 9);
unset($data);
$fp = fopen($path2.$to_dir.$filename.".zip", "xb+");
//$fp = fopen($path.$to_dir.$filename.".zip")
fwrite($fp, $gzdata);
unset($gzdata);
fclose($fp);
unlink($path.$from_dir.$filename);

?>

【问题讨论】：

@KIKOSoftware 感谢您的理解。编辑了第一篇文章和代码。

【参考方案1】：

看看rename()。

这是一个 PoC：

function move_files($src, $dst)

    $dh = opendir($src);
    if (!$dh) 
        return false;
    

    // Combine the letters before the first dash/hyphen (-),
    // and the letters after (and including) the first dot/period (.)
    // after the first dash/hyphen (-).
    $regex = '/^(.+?)-(?:.+?)(\..+?\.zip)$/';
    $moved = 0;
    $total = 0;

    while (($filename = readdir($dh)) !== false) 
        if (filetype("$src$filename") !== 'file') 
            continue;
        

        if (!preg_match($regex, $filename)) 
            continue;
        

        $total++;
        $new_filename = preg_replace($regex, "$1$2", $filename);
        $moved += (int)rename($src, "$dst$new_filename");
    

    closedir($dh);

    return [$moved, $total];


$srcDir = '/src/';
$dstDir = '/dst/';
$res = move_files($src, $dst);

if (!$res) 
    // Error


list($moved, $total) = $res;
var_dump($moved, $total);

【讨论】：

@loydb 感谢您的回答。我收到错误：Invalid argument in /root/arch.php on line 110。第 110 行如下所示：$moved += (int)rename($src, "$dst$new_filename");

@Erikas 你能转储rename() 之前的变量吗？即var_dump($src, $dst, $filename, $new_filename);.