使用AS3和PHP錯誤將文件上傳到服務器

Question

我正嘗試使用AS3和PHP將文件上傳到服務器。這是我的AS3代碼，然後是PHP代碼。我想要上傳到的文件夾是可寫的。和文件大小約20Kb。 php腳本在我的服務器上，並且flash文件調用它。

var UPLOAD_URL: String ="linktophpscriptonMysite" 
var fr: FileReference; 
var request: URLRequest = new URLRequest(); 
request.url = UPLOAD_URL; 

function startThis(): void { 

    fr = new FileReference(); 
    fr.addEventListener(Event.SELECT, selectHandler); 
    fr.addEventListener(Event.OPEN, openHandler); 
    fr.addEventListener(ProgressEvent.PROGRESS, progressHandler); 
    fr.addEventListener(Event.COMPLETE, completeHandler); 
    fr.addEventListener(IOErrorEvent.IO_ERROR, ioErrorHandler); 
    startUpload() 
} 

function startUpload(): void { 
    try { 
     var success: Boolean = fr.browse(); 
     trace("success") 
    } catch (error: Error) { 
     trace("Unable to browse for files.", Error); 
    } 
} 

function progressHandler(event: ProgressEvent): void { 
    trace(event.bytesLoaded, event.bytesTotal); 

} 
function ioErrorHandler(event: IOErrorEvent): void { 
    //trace("Some error ", event.target.data.systemResult); 
    //systemResult is echoed by PHP 

} 
function openHandler(event: Event): void { 
    try { 
     //var success: Boolean = fr.browse(); 
    } catch (error: Error) { 
     trace("Unable to browse for files.", Error); 
    } 

} 
function completeHandler(event: Event): void { 
    trace(event.target.data.systemResult); 
//this reads the result, again, from PHP echo "systemResult=all is good"; 


} 

function selectHandler(event: Event): void { 

    fr.upload(request); 
} 

然後，這裏是PHP代碼：此代碼是一般的上傳腳本我在PHP手冊網站上發現

<?php 

header('Content-Type: text/plain; charset=utf-8'); 

try { 


    // Undefined | Multiple Files | $_FILES Corruption Attack 
    // If this request falls under any of them, treat it invalid. 
    if (
     !isset($_FILES['upfile']['error']) || 
     is_array($_FILES['upfile']['error']) 
    ) { 
     echo "systemResult=Error"; 
     throw new RuntimeException('Invalid parameters.'); 

    } 

    // Check $_FILES['upfile']['error'] value. 
    switch ($_FILES['upfile']['error']) { 
     case UPLOAD_ERR_OK: 
      break; 
     case UPLOAD_ERR_NO_FILE: 
      throw new RuntimeException('No file sent.'); 

     case UPLOAD_ERR_INI_SIZE: 
     case UPLOAD_ERR_FORM_SIZE: 
      throw new RuntimeException('Exceeded filesize limit.'); 
     default: 
      throw new RuntimeException('Unknown errors.'); 
    } 

    // You should also check filesize here. max is 100 mb 
    if ($_FILES['upfile']['size'] > 10000000) { 
     throw new RuntimeException('Exceeded filesize limit.'); 
    } 

    // DO NOT TRUST $_FILES['upfile']['mime'] VALUE !! 
    // Check MIME Type by yourself. 
    $finfo = new finfo(FILEINFO_MIME_TYPE); 
    if (false === $ext = array_search(
     $finfo->file($_FILES['upfile']['tmp_name']), 
     array(
      'jpg' => 'image/jpeg', 
      'png' => 'image/png', 
      'gif' => 'image/gif', 
     ), 
     true 
    )) { 
     throw new RuntimeException('Invalid file format.'); 
    } 

    // You should name it uniquely. 
    // DO NOT USE $_FILES['upfile']['name'] WITHOUT ANY VALIDATION !! 
    // On this example, obtain safe unique name from its binary data. 
    if (!move_uploaded_file(
     $_FILES['upfile']['tmp_name'], 
     sprintf('./uploads/%s.%s', 
      sha1_file($_FILES['upfile']['tmp_name']), 
      $ext 
     ) 
    )) { 
     throw new RuntimeException('Failed to move uploaded file.'); 
    } 

    echo 'File is uploaded successfully.'; 

} catch (RuntimeException $e) { 

    echo $e->getMessage(); 

} 

?> 

我遇到的問題是，該文件沒有得到上傳，我沒有從PHP獲得任何反饋，爲什麼。

感謝您的幫助

UPDATE： 謝謝@akmozo的答覆和答覆。就像我在我的評論說，這個劇本的工作

<?php 

$uploads_dir = './uploads/'; 

    if($_FILES['Filedata']['error'] == 0){ 

    if(move_uploaded_file($_FILES['Filedata']['tmp_name'],  $uploads_dir.$_FILES['Filedata']['name'])){ 

    echo 'ok'; 
    echo 'systemResult=Awesome'; 

    exit(); 

    } 

    } 

    echo 'error'; 
    echo 'systemResult=did not work'; 

    exit(); 

    ?> 

Answer 1

默認情況下，FileReference對象的上傳數據字段名稱是"Filedata"，這就是你應該在你的PHP代碼（$_FILES['Filedata'] ...）用什麼。

您可以在FileReference.upload()功能的過程中更改名稱：

fr.upload(request, 'upfile'); 

希望可以幫助