AH00526 – apache 2.4需要ip范围

我正在使用FreeBSD 9.2-RELEASE-p5 w / apache24-2.4.12包:

 [root@j ~]# uname -a FreeBSD X 9.2-RELEASE-p5 FreeBSD 9.2-RELEASE-p5 #0 r265974: Wed May 14 07:57:04 PDT 2014 root@X:/usr/obj/usr/src/sys/R610 amd64 [root@j ~]# httpd -v Server version: Apache/2.4.12 (FreeBSD) Server built: Feb 4 2015 19:03:59 [root@j ~]# 

我想阻止来自我的Apache的某个networking(118.168.0.0 – 118.171.255.255),使用以下configuration:

 <RequireAll> Require not ip 118.168-171.0.0/16 Require all granted </RequireAll> 

而我正在运行一个语法检查,我得到如下:

 [root@j /usr/local/etc/apache24/Includes]# httpd -t AH00526: Syntax error on line 4 of /usr/local/etc/apache24/Includes/_RequireAll.: ip address '118.168-171.0.0/16' appears to be invalid [root@j /usr/local/etc/apache24/Includes]# 

如果我不能使用whois中定义的范围,那么如何正确确定子网掩码,这样就可以在Apache的httpd中使用了? 或者我写错了,因为我正在使用范围和子网掩码,并有适当的方式来定义范围为此。

*更新*

 [root@j /usr/local/etc/apache24/Includes]# grep 118 _RequireAll. Require not ip 118.168 118.169 118.170 118.171 [root@j /usr/local/etc/apache24/Includes]# httpd -t Syntax OK [root@j /usr/local/etc/apache24/Includes]# 

尝试以下

 Require not ip 118.168 118.169 118.170 118.171