Question

我编写了一个脚本来上传文件并将路径存储在数据库表中，以便下载。我有以下代码：

 <?php require("includes.php");
 ?><!DOCTYPE html>
 <html>
 <head>
 </head>
 <body>
 <?php if(isset($_FILES["upload"])==TRUE)
 {
      $errors           = array();
      $excluded      = array("exe", "zip", "js", "msi");
      /* If the contents of the file are to be held in the database then checking the extension is somewhat unneccessary but, hey, lets get rid of the files we know we don't want and then check the mime type. */
      $name               = $_FILES["upload"]["name"];
      $size               = $_FILES["upload"]["size"];     
      $type               = $_FILES["upload"]["type"];
      $temp               = $_FILES["upload"]["tmp_name"];     
      $extension = explode(".", $name);
      $extension = end($extension);
      if(in_array($extension, $excluded)==TRUE)
      {
           $errors[] ="This file may not be uploaded";
      }
      if(empty($errors)==FALSE)
      {
           foreach($errors as $error)
           {
                echo "<p>{$error}</p>\n";
           }
      }
      else
      {
           $year = date("Y");
           $month = date("m");
           $day = date("d");
           $name = strtolower(str_replace(" ", "_", $name));
           $path = "uploads/{$day}-{$month}-{$year}";
           if(file_exists($path)==FALSE)
           {
                mkdir($path);
           }
           elseif(file_exists("{$path}/{$name}")==FALSE)
           {
                move_uploaded_file($temp, "{$path}/{$name}");
                $add = add_file_to_database($connection, "{$path}/{$name}");
                if($add[0]==TRUE)
                {
                     $url = "http://example.com/uploads.php?id={$add[1]}";
                     echo "<p>This file has been uploaded, it can be found at: <a href=\"{$url}\">{$url}</a></p>";
                }
                else
                {
                     echo "<p>I'm sorry but an error happened</p>";
                }
           }
      }
 }
 ?>
 <form action="index.php" method="post" enctype="multipart/form-data">
      <label for="upload">Upload a file: </label><input type="file" name="upload" id="upload"><br>
      <input type="submit" value="Upload" name="submit">
 </form>
 </body>
 </html>

uploads.php中的代码是：

 <?php require("includes.php");
 $file = get_uploaded_file($connection, $_GET["id"]);
 header("Content-Type:{$file[0]}");
 echo file_get_contents($file[1]);
 ?>

如果我上传了一个jpeg文件，一个PDF或txt文件，那么它会在浏览器中显示，但是如果我上传一个word文件或MP3，那么我希望它作为普通文件下载而不是uploads.php

不确定我将如何实现这一目标。你能给我一些关于如何做到这一点的想法，所以如果我上传“demo.mp3”并获得1的ID，那么我希望它下载一个名为“demo.mp3”的文件。只是认为MS Word无法识别自己的MIME类型

使用PHP上传文件：需要保留扩展名

0 个答案: