小编Jig*_*r D的帖子

我尝试使用<?php echo php_uname("m"); ?>,它返回i586,但我在Windows 7 64位,我可以在我的计算机属性中看到.所以我期待输出中的x86_64.有谁知道如何在PHP中确定OS体系结构？

对于Mac OS X我也想要同样的东西.任何帮助,将不胜感激.

一切都在我的Windows 7中完美运行.

问题是当我将domain1.com添加为VirtualHost时,localhost的DocumentRoot更改为VirtualHost的DocumentRoot.

例如:当我访问http:// localhost时,我打开了为domain1.com指定的DocumentRoot,而不是httpd.conf中指定的那个.

我的httpd-vhosts.conf文件是:

NameVirtualHost 127.0.0.1:80
NameVirtualHost domain1.com:80
<VirtualHost domain1.com:80>
<Directory "e:/program files/apache/htdocs/domain1.com">
    Options FollowSymLinks Indexes
    AllowOverride All
    Order deny,allow
    allow from All
</Directory>
ServerName domain1.com
ServerAlias domain1.com
ScriptAlias /cgi-bin/ "e:/program files/apache/htdocs/domain1.com/cgi-bin/"
DocumentRoot "e:/program files/apache/htdocs/domain1.com"
ErrorLog "E:/Program Files/apache/logs/domain1.com.err"
CustomLog "E:/Program Files/apache/logs/domain1.com.log" combined
</VirtualHost>

我的主机文件:

127.0.0.1   domain1.com

我的httpd.conf文件:

DocumentRoot "e:/program files/apache/htdocs"
<Directory />
    Options FollowSymLinks
    AllowOverride None
    Order deny,allow
    Deny from all
</Directory>
<Directory "e:/program files/apache/htdocs">
    Options Indexes FollowSymLinks
    AllowOverride All
    Order allow,deny
    Allow from …