14

I am using php script to provide download from my website after a requisite javascript timer this php script is included which causes the download. But the downloaded file is corrupt no matter whatever I try. Can anyone help me point out where am I going wrong.

This is my code

     <?php
include "db.php";    
 $id = htmlspecialchars($_GET['id']);
 $error = false;
    $conn = mysql_connect(DB_HOST,DB_USER,DB_PASSWORD);
    if(!($conn)) echo "Failed To Connect To The Database!";
    else{   
        if(mysql_select_db(DB_NAME,$conn)){
            $qry = "SELECT Link FROM downloads WHERE ID=$id";
            try{
                $result = mysql_query($qry);
                if(mysql_num_rows($result)==1){
                    while($rows = mysql_fetch_array($result)){
                        $f=$rows['Link'];
                    }
                    //pathinfo returns an array of information
                    $path = pathinfo($f);
                    //basename say the filename+extension
                    $n = $path['basename'];
                    //NOW comes the action, this statement would say that WHATEVER output given by the script is given in form of an octet-stream, or else to make it easy an application or downloadable
                    header('Content-type: application/octet-stream');
                    header('Content-Length: ' . filesize($f));
                    //This would be the one to rename the file
                    header('Content-Disposition: attachment; filename='.$n.'');
                    //Finally it reads the file and prepare the output
                    readfile($f);
                    exit();
                }else $error = true;
            }catch(Exception $e){
                $error = true;
            }
            if($error) 
            {
                header("Status: 404 Not Found");
                }
        }
  }
?>
4

3 回答 3

27

在打开更多输出缓冲区的情况下，这对我很有帮助。

//NOW comes the action, this statement would say that WHATEVER output given by the script is given in form of an octet-stream, or else to make it easy an application or downloadable
header('Content-type: application/octet-stream');
header('Content-Length: ' . filesize($f));
//This would be the one to rename the file
header('Content-Disposition: attachment; filename='.$n.'');
//clean all levels of output buffering
while (ob_get_level()) {
    ob_end_clean();
}
readfile($f);
exit();
于 2015-03-12T06:19:39.983 回答
14

首先，正如一些人在评论中指出的那样，删除<?php第一行开始 PHP 标记 ( ) 之前的所有空格，这样就可以解决问题（除非该文件包含在其他文件中或需要该文件）。

当您在屏幕上打印任何内容时，即使是一个空格，您的服务器也会发送标题以及要打印的内容（在这种情况下，您的空格）。为防止这种情况发生，您可以：

a) 在写完标题之前不要打印任何东西；

b) 运行 ob_start() 作为脚本中的第一件事，编写内容，编辑标题，然后在您希望将内容发送到用户浏览器时使用 ob_flush() 和 ob_clean()。

在 b) 中，即使您成功写入标头而没有出现错误，空格也会损坏您的二进制文件。你应该只写你的二进制内容，而不是二进制内容的几个空格。

ob_前缀代表输出缓冲区。调用 时ob_start()，您告诉您的应用程序您输出的所有内容（echo、printf等）都应该保存在内存中，直到您明确告诉它“去”（ob_flush()）给客户端。这样，您将输出与标题一起保存，当您完成编写它们时，它们将与内容一起发送。

于 2012-11-09T17:15:23.800 回答
2 
           ob_start();//add this to the beginning of your code 

      if (file_exists($filepath) && is_readable($filepath) ) {
header('Content-Description: File Transfer');
 header("Content-Type: application/octet-stream");
 header("Content-Disposition: attachment; filename=$files");
 header('Expires: 0');
header('Cache-Control: must-revalidate, post-check=0, pre-check=0');
   header('Pragma: public');
 header("content-length=".filesize($filepath));
 header("Content-Transfer-Encoding: binary");

   /*add while (ob_get_level()) {
       ob_end_clean();
         }  before readfile()*/
  while (ob_get_level()) {
ob_end_clean();
   }
    flush();
   readfile($filepath);
于 2020-03-25T19:56:28.970 回答