如何编写php子类将数据和文件插入数据库？

我正在使用它来尝试允许用户在ajax表单中上载文件。我在iframe中隔离了表单的文件部分，因为表单的一些JS与文件上传系统发生冲突。该系统现在可以工作了——只有一个关键的例外。这是一个漂亮的系统，但在我看来，它的能力有限，相当复杂。它成功地将一个文件保存到服务器磁盘上，但似乎我必须编写一个子类，使其将文件作为blob保存到mysql数据库中。几天来，我一直在努力让它发挥作用，但都没有成功。我对课程不熟悉——尽管我读了很多书（并没有像我想的那样理解）

这是我试图输入的子类的代码，也是文件上传脚本在提交后立即与之对话的代码

<?php
/*
 * jQuery File Upload Plugin PHP Example 5.14
 * https://github.com/blueimp/jQuery-File-Upload
 *
 * Copyright 2010, Sebastian Tschan
 * https://blueimp.net
 *
 * Licensed under the MIT license:
 * http://www.opensource.org/licenses/MIT
 *
 */
error_reporting(E_ALL | E_STRICT);
require('UploadHandler.php');

class MyUploadHandler extends UploadHandler {
    /*
    * Maintain our database:
    * For each uploaded file, add an entry to the tbl_xxx
    */
    protected function handle_file_upload($uploaded_file, $name, $size, $type, $error, $index = null, $content_range = null) {
        // this does what would have automatically been done before we subclasses anything
        $file = parent::handle_file_upload($uploaded_file, $name, $size, $type, $error, $index, $content_range);

        // Now we add some extra processing if the file uploaded with no problems
        if(empty($file->error)) {
            // include the funky stuff

            // Grab the includes
            include '../conf/Funcs.php';
            include '../conf/DBconfig.php';

            // Grab critical data
            $appID = $_POST['appID'];
            $uaID = $_POST['uaID'];
            $uploaders_uID = $_POST['uID'];
            $applicationKey = $_POST['applicationKey'];
            $token = $_POST['token'];
            $dateUploaded = time();

            // Define our arrays which we will be using
            $fileName = array();
            $fileMime = array();
            $fileSize = array();
            $fileData = array();

            // And grab our file data
            $fileName = $_FILES['files']['name'];
            $fileMime = $_FILES['files']['type'];
            $fileSize = $_FILES['files']['size'];
//          $fileData = fopen($_FILES['files']['tmp_name'], "rb");
            $fileData = $_FILES['files']['files[]'];



            foreach ($fileName as $nameitem) {
                foreach ($fileMime as $mimeitem) {
                    foreach ($fileSize as $sizeitem) {
//                      foreach ($fileData as $dataitem) {
                            echo "$nameitem\n";
                            echo "$mimeitem\n";
                            echo "$sizeitem\n";
//                          echo "$dataitem\n";
//           THIS IS WHERE THE DB COMMUNICATION CHUNK
//           WOULD BE GOING IF THIS WORKED.


//                      }
                    }
                }
            }


        }

    return $file;
    }
} // end of class

$upload_handler = new MyUploadHandler($site_params);

// THIS IS THE DB COMMUNICATION PART THAT WILL GO
// AS MARKED UP ABOVE, ONCE THE THING WORKS
//  try {
//      $stmt = $conn->prepare("INSERT INTO $database.appsFiles (`appID`, `uaID`, `uploaders_uID`, `dateUploaded`, `applicationKey`, `fileName`, `fileMime`, `fileSize`, `file`)
//      VALUES
//      (?,?,?,?,?,?,?,?,?)");
        // Bind Values
//      $stmt->bindParam(1, $appID, PDO::PARAM_INT, 11);
//      $stmt->bindParam(2, $uaID, PDO::PARAM_INT, 11);
//      $stmt->bindParam(3, $uploaders_uID, PDO::PARAM_INT, 11);
//      $stmt->bindParam(4, time(), PDO::PARAM_INT, 11);
//      $stmt->bindParam(5, $applicationKey, PDO::PARAM_STR, 8);
//      $stmt->bindParam(6, $fileName, PDO::PARAM_STR, 255);
//      $stmt->bindParam(7, $fileMime, PDO::PARAM_STR, 128);
//      $stmt->bindParam(8, $fileSize, PDO::PARAM_INT, 20);
//      $stmt->bindParam(9, $file, PDO::PARAM_LOB, 5120000);

        // Execute the query
//      $stmt->execute();
//  } catch(PDOException $e) { catchMySQLerror($e->getMessage()); }
?>

我知道目前没有数据检查。当我解决了如何将包含所有相关信息的文件放入数据库时，我会在后面添加这些信息

---

非常感谢您的帮助。

问题是什么？另外，您确定应该将文件存储在数据库中吗？也许您应该将文件存储在文件系统中，并将文件路径存储在数据库中？问题是如何将文件放入数据库中。我的课堂努力有什么问题？是的，它们需要在数据库中，而不是凌乱服务器文件系统。