在php中将文件从文件夹复制到文件夹树[关闭]答案

【问题标题】：Copy Files from folders to folder tree in php [closed]在php中将文件从文件夹复制到文件夹树[关闭]
【发布时间】：2013-05-08 19:47:19
【问题描述】：

我有大量书籍需要按语言和图书馆 ID 进行分类。

文件名的结构是这样的，并分布在许多文件夹中：

ENG_FILENAME_LIBRARYNAME1.pdf
ENG_FILENAME_LIBRARYNAME2.pdf
SPA_FILENAME_LIBRARYNAME1.pdf
SPA_FILENAME_LIBRARYNAME2.pdf

我需要将它们移动到这样的文件夹中

ENG
->LIBRARYNAME1
--ENG_FILENAME_LIBRARYNAME1.pdf
->LIBRARYNAME2
--ENG_FILENAME_LIBRARYNAME2.pdf

这是我的代码：

foreach (glob("C:/wamp/www/projects/filemove/eth/*") as $folderpath) {
$foldername = preg_replace('/.*?\/(.*?)/', '$1', $folderpath);
    foreach (glob("C:/wamp/www/projects/filemove/eth/*/*") as $librarypath) {
    $libraryname = preg_replace('/.*?\/(.*?)/', '$1', $librarypath);

        foreach (glob("C:/wamp/www/projects/filemove/pdf/*.pdf") as $filepath) {
           $ethologue = preg_replace('/(.*?)_.*/', '$1', $filepath);
           $library = preg_replace('/.*?_.*?_.*?_(.*?)_.*/', '$1', $filepath);
           $filename = preg_replace('/.*?\/(.*?)/', '$1', $filepath);

           if ($ethologue = $foldername ) {
           if ($library = $libraryname) {
                copy($filepath, $librarypath);
                }
           }
        }
    }
}

提前致谢！

【问题讨论】：

你忘了你的问题？
啊，我想我的问题归结为如何修复我相当粗糙的代码。

标签： php foreach glob organization file-management

【解决方案1】：

你可以试试这个

define("FILE_COPY", 1);
define("FILE_MOVE", 2);

$dirSource = __DIR__ . "/case";
$dirDestination = $dirSource . "/books";
$cache = $errors = array();
$allowDuplicate = false;
$mode = FILE_COPY; // copy or move file ;

try {
    $dirIT = new FilesystemIterator($dirSource, FilesystemIterator::SKIP_DOTS);
    $regexIT = new RegexIterator($dirIT, '/.pdf$/i');

    // Make Directory
    if (! is_dir($dirDestination) and ! @mkdir($dirDestination, 0777, true)) {
        throw new Exception("Destination Folder ($dirDestination) does not exist");
    }

    foreach($regexIT as $splFile) {
        $hash = md5_file($splFile);
        $ext = "." . $splFile->getExtension();

        // Don't take duplicates
        if (! $allowDuplicate && isset($cache[$hash])) {
            continue;
        }
        // Split File Name
        list($category, $name, $subCategory) = explode("_", $splFile->getBasename($ext));
        if (empty($category) || empty($name) || empty($subCategory)) {
            $errors[] = "File ($splFile) does not have valid name format";
        }

        // New File path
        $path = sprintf("%s/%s/%s", $dirDestination, $category, $subCategory);
        if (! is_dir($path) and ! @mkdir($dirDestination, 0777, true)) {
            throw new Exception("Destination Folder ($path) does not exist");
        }

        $fileTmp = $path . DIRECTORY_SEPARATOR . $splFile->getFileName();

        if (is_file($fileTmp)) {

            if (! $allowDuplicate) {
                // Check if file is duplicate
                $copyhash = md5_file($fileTmp);
                if ($hash == $copyhash) {
                    continue;
                }
            }
            $x = 1;
            while(is_file($fileTmp)) {
                $fileTmp = $path . DIRECTORY_SEPARATOR . $splFile->getBasename($ext) . "_" . $x . $ext;
                $x ++;
            }
        }

        if ($mode == FILE_COPY) {
            if (! copy($splFile, $fileTmp))
                $errors[] = "File ($splFile) Failed Copy";
        } else {
            if (! rename($splFile, $fileTmp))
                $errors[] = "File ($splFile) Failed Move";
        }
        $cache[$hash] = true;
    }
} catch (Exception $e) {
    echo $e->getMessage();
}

【讨论】：

【解决方案2】：

我立即看到的唯一一件事是您正在 if 语句中进行赋值。这是故意的吗？

if ($ethologue = $foldername ) {
   if ($library = $libraryname) {

这应该是一个比较：

if ($ethologue == $foldername ) {
   if ($library == $libraryname) {

【讨论】：