我已经在 Google 上搜索并研究该主题很长时间了,但只找到一份详尽的指南......但那没有用。
我想在-上运行PHP
mod_fcgid 。Apache2
CentOS 5.3
使用我当前的配置,我得到了500 Internal Server Error
-files .php
。你能帮我找出原因吗?:)
httpd.conf:
<Directory "/var/www/html/">
Options Indexes ExecCGI
AllowOverride None
Allow from all
AddHandler fcgid-script .php
FCGIWrapper /var/www/cgi-bin/php5/php-fcgi-starter .php
</Directory>
在 /var/www/cgi-bin/php5/php-fcgi-启动器:
#!/bin/sh
PHPRC=/etc/
export PHPRC
export PHP_FCGI_MAX_REQUESTS=5000
export PHP_FCGI_CHILDREN=8
exec /usr/bin/php-cgi
php-cgi -v:
PHP 5.2.6 (cgi-fcgi) (built: May 2 2008 16:01:17)
Copyright (c) 1997-2008 The PHP Group
Zend Engine v2.2.0, Copyright (c) 1998-2008 Zend Technologies
with the ionCube PHP Loader v3.1.29, Copyright (c) 2002-2007, by ionCube Ltd.
SeLinux
已关闭。
编辑:日志…
[Mon Nov 02 05:17:49 2009] [notice] Apache/2.2.3 (CentOS) configured -- resuming normal operations
[Mon Nov 02 05:17:53 2009] [error] Not even headers for me :(
[Mon Nov 02 05:17:54 2009] [notice] mod_fcgid: call /var/www/html/phpinfo.php with wrapper /var/www/cgi-bin/php5/php-fcgi-starter
[Mon Nov 02 05:17:54 2009] [notice] mod_fcgid: server /var/www/html/phpinfo.php(13917) started
[Mon Nov 02 05:17:57 2009] [notice] mod_fcgid: process /var/www/html/phpinfo.php(13917) exit(communication error), terminated by calling exit(), return code: 120
SuExec_log:
[2009-11-02 04:55:27]: uid: (10001/www) gid: (2524/2524) cmd: php5
[2009-11-02 04:55:27]: target uid/gid (10001/2524 or 2523) mismatch with directory (10001/2523) or program (0/0)
[2009-11-02 04:57:19]: uid: (10001/www) gid: (2524/2524) cmd: php-fcgi-starter
[2009-11-02 04:57:19]: target uid/gid (10001/2524 or 2523) mismatch with directory (0/0) or program (0/0)
谢谢 :)
答案1
SUExec 要求目录 ( /var/www/html/
) 和二进制文件 ( /var/www/cgi-bin/php5/php-fcgi-starter
) 由同一用户/组拥有。看起来10001/2523
拥有目录,而 root 拥有 php-fcgi-starter。修复该问题,或关闭 SUExec。