将'用户名'保存为文件名

Question

我想知道是否有人可以帮助我。

我使用Aurigma的Image Uploader，为了保存上传的图像，我把这个脚本放在一起。

<?php 

//This variable specifies relative path to the folder, where the gallery with uploaded files is located. 
//Do not forget about the slash in the end of the folder name. 
$galleryPath = 'UploadedFiles/'; 

require_once 'Includes/gallery_helper.php'; 

require_once 'ImageUploaderPHP/UploadHandler.class.php'; 

/** 
* FileUploaded callback function 
* @param $uploadedFile UploadedFile 
*/ 
function onFileUploaded($uploadedFile) { 

    $packageFields = $uploadedFile->getPackage()->getPackageFields(); 
    $userid = $packageFields["userid"]; 
    $locationid= $packageFields["locationid"]; 

    global $galleryPath; 

    $absGalleryPath = realpath($galleryPath) . DIRECTORY_SEPARATOR; 
    $absThumbnailsPath = $absGalleryPath . 'Thumbnails' . DIRECTORY_SEPARATOR; 

    if ($uploadedFile->getPackage()->getPackageIndex() == 0 && $uploadedFile->getIndex() == 0) { 
    initGallery($absGalleryPath, $absThumbnailsPath, FALSE); 
    } 

    $dirName = $_POST['folder']; 
    $dirName = preg_replace('/[^a-z0-9_\-\.()\[\]{}]/i', '_', $dirName); 
    if (!is_dir($absGalleryPath . $dirName)) { 
    mkdir($absGalleryPath . $dirName, 0777); 
    } 

    $path = rtrim($dirName, '/\\') . '/'; 

    $originalFileName = $uploadedFile->getSourceName(); 

    $files = $uploadedFile->getConvertedFiles(); 

    // save converter 1 

    $sourceFileName = getSafeFileName($absGalleryPath, $originalFileName); 
    $sourceFile = $files[0]; 
    /* @var $sourceFile ConvertedFile */ 
    if ($sourceFile) { 
    $sourceFile->moveTo($absGalleryPath . $sourceFileName); 
    } 

    // save converter 2 

    $thumbnailFileName = getSafeFileName($absThumbnailsPath, $originalFileName); 
    $thumbnailFile = $files[1]; 
    /* @var $thumbnailFile ConvertedFile */ 
    if ($thumbnailFile) { 
    $thumbnailFile->moveTo($absThumbnailsPath . $thumbnailFileName); 
    } 

    //Load XML file which will keep information about files (image dimensions, description, etc). 
    //XML is used solely for brevity. In real-life application most likely you will use database instead. 
    $descriptions = new DOMDocument('1.0', 'utf-8'); 
    $descriptions->load($absGalleryPath . 'files.xml'); 

    //Save file info. 
    $xmlFile = $descriptions->createElement('file'); 
    $xmlFile->setAttribute('name', $_POST['folder'] . '/' . $originalFileName); 
    $xmlFile->setAttribute('source', $sourceFileName); 
    $xmlFile->setAttribute('size', $uploadedFile->getSourceSize()); 
    $xmlFile->setAttribute('originalname', $originalFileName); 
    $xmlFile->setAttribute('thumbnail', $thumbnailFileName); 
    $xmlFile->setAttribute('description', $uploadedFile->getDescription()); 
    //Add additional fields 
    $xmlFile->setAttribute('userid', $userid); 
    $xmlFile->setAttribute('locationid', $locationid); 
    $xmlFile->setAttribute('folder', $dirName); 
    $descriptions->documentElement->appendChild($xmlFile); 
    $descriptions->save($absGalleryPath . 'files.xml'); 
} 

$uh = new UploadHandler(); 
$uh->setFileUploadedCallback('onFileUploaded'); 
$uh->processRequest(); 
?> 

我想要做的是替换文件名的files元素，并与username替换它，所以每个保存文件夹和相关文件可以中鉴定到每个用户。

我添加了一个用户名文本字段到这个脚本保存来自

我觉得我说得很对，这是一个需要改变$descriptions->save($absGalleryPath . 'files.xml');线的形式。

因此，我尝试将此更改为$descriptions->save($absGalleryPath . '$username.xml，$descriptions->save($absGalleryPath . $username '.xml，但这些都没有奏效，因此我不太清楚需要更改哪些内容。

我只是想知道是否有人可以看看这个请让我知道我要去哪里错了。

非常感谢

Answer 1

'$ username.xml' 将被解释为$ username.xml，你需要使用 “$ username.xml”。单引号“禁用”字符串内部的变量使用。

你正在尝试可能是一个坏主意，因为你正在创建一个用户名不能包含像“/”这样的特殊字符。如果你有一条规则阻止“/”成为用户名的一部分，也许不是问题。