PHP readfile（）导致文件下载损坏

Question

[我正在使用php脚本在需要的JavaScript计时器之后提供从我的网站提供的下载，此php脚本已包括在内，这导致下载。但是无论我如何尝试，下载的文件都已损坏。谁能帮我指出我要去哪里了。

这是我的代码

     <?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");
                }
        }
  }
?> 

Answer 1

首先，正如某些人在评论中指出的那样，删除第一行中打开PHP标记（<?php）之前的所有空格，这应该可以解决问题（除非此文件包含在其他文件中或由其他文件要求） ）。

当您在屏幕上打印任何内容时，即使只有一个空格，您的服务器也会将标题和要打印的内容一起发送（在这种情况下为空白）。为了防止这种情况发生，您可以：

a）在完成标题之前不打印任何内容；

b）首先将ob_start（）作为脚本中的第一件事，编写内容，编辑标题，然后在希望将内容发送到用户的浏览器时进行ob_flush（）和ob_clean（）的处理。

在b）中，即使您成功编写了头文件而没有收到错误，空格也会破坏您的二进制文件。您只应编写二进制内容，而不能在二进制内容中写一些空格。

ob _前缀代表输出缓冲区。调用ob_start()时，您告诉应用程序输出的所有内容（echo，printf等）都应保留在内存中，直到您明确告诉它“去”（ob_flush()）给客户端为止。这样，您就可以将输出与标头一起保存，当您完成对标头的编写后，它们将与内容一起发送。

Answer 2

[这在打开更多输出缓冲区的情况下对我有帮助。

//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();

Answer 3

           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);