PHP:将视频和图像上传到同一文件夹。

时间:2014-04-16 04:55:08

标签: php html mysql

我正在完成一个简单的视频文件上传。但我现在也要将图像上传到同一个文件夹。可能吗??

这是我的代码:

newvideo.html

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
         <html xmlns="http://www.w3.org/1999/xhtml">
       <head>
      <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
             <title>Untitled Document</title>

       </head>
          <body>
            <form enctype="multipart/form-data" action="upload_file.php" method="POST">
             Video Name :<input type="text" name="vname"/><br />
                 Category : <input type="text" name="cat"/><br />
                 Keywords : <input type="text" name="kwd"/><br />
                     Please choose a file: <input name="uploaded" type="file" /><br />
                  Please choose a image:<input type="file" name="image"><br />
                            <input type="submit" value="Upload" />
                               </form> 
            </body>
          </html>

upload_file.php

<?php 
    include "config.php";
   $videoname = $_REQUEST["vname"];
 $category = $_REQUEST["cat"];
 $keyword = $_REQUEST["kwd"];
 $target = "C:/xampp/htdocs/upload/"; 
  $target = $target . basename( $_FILES['uploaded']['name']) ; 
   $ok=1; 
    if ($_FILES["uploaded"]["error"] > 0)
    {
   echo "Apologies, an error has occurred.";
   echo "Error Code: " . $_FILES["uploaded"]["error"];
    }
   if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)) 
   {
  echo "The file ". basename( $_FILES['uploaded']['name']). " has been uploaded";
     } 
       else 
           {
         echo "Sorry, there was a problem uploading your file.";
        }
             ?>
               <?php
                  $path = str_replace("C:/xampp/htdocs","http://localhost",$target);
               $query = "INSERT INTO videos(videoname, category, keywords, path)";
               $query .= "VALUE ('$videoname', '$category', '$keyword', '$path')";
                $result = mysql_query($query);
                ?>

0 个答案:

没有答案