我一直在谷歌search和研究的话题相当长一段时间，但只find一个彻底的指导…这是行不通的。
我想通过Apache2 mod_fcgid运行PHP – CentOS 5.3 。
与我目前的configuration，我得到500 Internal Server Error .php 。 你能帮我弄清楚为什么？ 🙂
httpd.conf中：
<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目录/ PHP5 / PHP-fcgi的起动器：
#!/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被打开了。
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 ：）
SUExec要求目录（ /var/www/html/ ）和二进制文件（ /var/www/cgi-bin/php5/php-fcgi-starter ）由同一个用户/组拥有。 看起来像10001/2523拥有目录，而根拥有php-fcgi-starter。 修复，或closuresSUExec。