Question

我有以下PHP代码

$filename = 'a56.flv';
$file = "C:/xampp/htdocs/site/flv/a56.flv";
header("Content-Type: application/force-download");
header('Content-Type: video/x-flv');
header('Content-Disposition: attachment; filename="' . addslashes($filename) . '"');
//header("Content-Type: application/octet-stream");
header("Content-Type: application/download; charset=UTF-8");
header("Content-Description: File Transfer");
header("X-Sendfile: $file");

当我直接调用脚本时（http：//localhost/site/get2.php？ftype = flv＆amp; filename = a56.flv）它工作正常但是只要调用（http：// localhost / site /） vsrc_a56.flv）它返回404页面和这个apache错误：

[Sun Jan 15 02:01:44 2012] [error] [client 127.0.0.1] (20024)The given path is misformatted or contained invalid characters: xsendfile: unable to find file: flv/a56.flv

我的.htaccess文件如下所示：

IndexIgnore .htaccess */.??* *~ *# */HEADER* */README* */_vti*
AddDefaultCharset UTF-8
#########ErrorDocument 404 
ErrorDocument 404 /site/404.php
RewriteEngine On
XSendFile on
RewriteRule ^vsrc_(.*)$  ./get2.php?ftype=flv&filename=$1 [L]

Answer 1

尝试在RewriteRule

之前添加RewriteBase /site/

RewriteEngine和X SendFile之间的冲突

1 个答案: