需要帮助处理XML HTTP文件上传请求

Question

因此，我查找了a tutorial上传文件并使用XML HTTP请求将文件发送到服务器。我跟着教程，但是，我认为我必须错过一些东西。当文件似乎被上传和发送时，“处理器”文件中的任何内容都不会被访问。有没有我需要编写的PHP函数来处理它？对于上下文，这里是我写的：

$(document).ready(function() 
{ 
    $('#upload-button').click(function(event) 
    { 
     $('#upload-button').removeClass("btn-danger"); 
    }); 

    $("#report-form").submit(function(event) 
    { 

     var form = document.getElementById('report-form'); 
     var fileSelect = document.getElementById('file-select'); 
     var uploadButton = document.getElementById('upload-button'); 

     event.preventDefault(); // Stop the event from sending the way it usually does. 

     uploadButton.value = 'Submitting...'; // Change text. 
     var files = fileSelect.files; 
     var maxfiles = <?php echo $config['Report_MaxFiles'] ?>; 
     var mfs = <?php echo $config['Report_MaxFileSize'] ?>; 
     if(files.length > maxfiles) // Make sure it's not uploading too many. 
     { 
      uploadButton.value = 'You uploaded too many files. The limit is ' + maxfiles + '.'; // Update button text. 
      $('#upload-button').addClass('btn-danger'); // Make the button red, if so. 
      return; 
     } 
     var formData = new FormData(); // Make a "form data" variable. 
     for (var i = 0; i < files.length; i++) { 
      var file = files[i]; 

      // Add the file to the request. 
      if(file.size/1000 > mfs) 
      { 
       uploadButton.value = 'One of the files is too big. The file size limit is ' + (mfs) + 'kb (' + (mfs/1000) + 'mb).'; 
       $('#upload-button').addClass('btn-danger'); 
       return; 
      } 
      formData.append('files[]', file, file.name); // Not really sure what this does, to be honest, 
                  // but I think it makes a file array. 
     } 

     var xhr = new XMLHttpRequest(); // Construct an XML HTTP Request 
     xhr.open('POST', 'assets/class/FileHandler.php', true); // Open a connection with my handler PHP file. 

     xhr.onload = function() 
     { 
      if (xhr.status === 200) 
      { 
       uploadButton.value = 'Files Submitted!'; // NOTE: I do get this message. 
      } 
      else 
      { 
       uploadButton.value = 'An error occurred.'; 
       $('#upload-button').addClass("btn-danger"); 
      } 
     }; 

     xhr.send(formData); // I think this is where it dies. 
    }); 
}); 

在“发送（FORMDATA）”行，我实际上并不知道它的发送。我是否在FileHandler.php中设置了某种侦听器，并在通过XML HTTP请求发送文件时激活它们？或者更具体地说，如何使用我的FileHandler.php文件将上传的文件保存到服务器？

编辑：我一直没能拿出在FileHandler.php文件比这个，我认为可能被称为当发送的任何其它的PHP代码（但它不是）：

编辑2：好吧，现在我有东西，但它不工作（没想到它）。我想我可能是错的使用变量：

<?php 

$uploaddir = 'data/reports/uploads/' . $_POST['id'] . "/"; 
$uploadfile = $uploaddir . basename($_FILES['files']['name']); 
echo "<script>console.log('RECEIVED');</script>"; 
echo '<pre>'; 
if (move_uploaded_file($_FILES['files']['tmp_name'], $uploadfile)) { 
    echo "File is valid, and was successfully uploaded.\n"; 
} else { 
    echo "Possible file upload attack!\n"; 
} 

echo 'Here is some more debugging info:'; 
print_r($_FILES); 

print "</pre>"; 

它不保存文件的目录，也不是打印文字讯息。我如何让我的report.php文件在FileHandler.php中执行这些事情？

Answer 1

感谢@FlorianLefèvre的帮助和耐心，我把它修好了。 :)

问题是与路径。它没有找到文件夹data/uploads /的路径，也没有找到目录。下面是我做的：

$uploaddir = '../../data/reports/uploads/' . $_POST['id'] . "/"; 
echo "NAME: " . $_FILES['files']['name'][0] . "\n"; 

foreach($_FILES['files']['name'] as $filenumber => $filename) 
{ 
    $uploadfile = $uploaddir . basename ($filename); 

    echo "UploadDir " . $uploaddir . "\n"; 
    echo "UploadFile " . $uploadfile . "\n"; 
    echo '<pre>'; 

    echo "MKDir for UploadDir which is: ". $uploaddir . "\n"; 
    mkdir ($uploaddir); 

    if (move_uploaded_file ($_FILES['files']['tmp_name'][$filenumber], $uploadfile)) 
    { 
     echo "File is valid, and was successfully uploaded.\n"; 
    } 
    else 
    { 
     echo "Possible file upload attack!\n"; 
    } 

    echo 'Here is some more debugging info:'; 

    print "</pre>"; 
} 

var_dump ($_FILES); 

我还没有摆脱的一些调试的东西还没有，但是这是通用的解决方案。