Apache+tomcat

Question

系统是用ssh开发的，其中当用户通过Apache访问跳转到tomcat，struts取到的ip值永远是127.0.0.1，直接通过：8080端口访问取到的ip值是正确的，
apache 中http.conf配置

[code="java"]ProxyPass /images !
ProxyPass /css !
ProxyPass /js !
ProxyPass /cms http://localhost:8080/cms
[/code]

另外还试过跳转
[code="java"]

ProxyPass http://localhost:8080/cms
ProxyPassReverse http://localhost:8080/cms

[/code]

jk没试过，看了网上配置都出错较郁闷，请问大家是怎么处理的

Answer 1

我也是用apache+tomcat的运行环间,我是用下边的代码取的客户端IP,没发现有错的:)
/**
* 如果通过了多级反向代理的话，X-Forwarded-For的值并不止一个，
* 而是一串Ｉｐ值，究竟哪个才是真正的用户端的真实IP呢？答案是取X-Forwarded-For
* 中第一个非unknown的有效IP字符串。如：X- Forwarded-For：192.168.1.110,
* 192.168.1.120, 192.168.1.130, 192.168.1.100用户真实IP为： 192.168.1.110
*
* @param request
* @return
*/
public static String getIpAddr(HttpServletRequest request) {
String ip = request.getHeader("x-forwarded-for");
if (ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
ip = request.getHeader("Proxy-Client-IP");
}
if (ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
ip = request.getHeader("WL-Proxy-Client-IP");
}
if (ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
ip = request.getRemoteAddr();
}
return ip;
}