2012-01-15 46 views
0
之间Confilict

我有以下的PHP代码RewriteEngine叙述和XSendFile

$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(HTTP://localhost/site/get2.php FTYPE = FLV &名= a56.flv)它工作正常,但只要用(HTTP://localhost/site/vsrc_a56.flv)调用返回404页,这Apache的错误:

[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 

凡我.htaccess文件就像如下:

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] 

回答

1

尝试重写规则之前添加RewriteBase /site/

+0

它只有R标志的RewriteRule工作^ _ VSRC(。*)$ ./get2.php?ftype=flv&filename=$1 [R,L] – Huseyin 2012-01-15 01:53:30

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