如果使用"localhost"URL,则无法从HttpServletRequest获取IP

Question

我在尝试从HttpServletRequest获取IP时出现问题,请先查看我的编码:

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.getHeader("HTTP_CLIENT_IP");
        }
        if (ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
            ip = request.getHeader("HTTP_X_FORWARDED_FOR");
        }
        if (ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
            ip = request.getRemoteAddr();
        }
        return ip; 

我的问题是如果打开的应用程序有以下URL(我的PC的URL是18.111,服务器部署在localhost上)"https://192.168.18.111:8443/test/main.html",我可以用上面的代码得到正确的URL,但是如果使用"https:// localhost:8443/test/main.html"打开,它将返回类似"0.1.0.1 ...."的函数,上面的函数,为什么这个函数不适用于"localhost"或者是谁知道有没有更好的方法从HttpServletRequest获取IP？

Answer 1

你方法的结果绝对正确.我假设你得到的数字是0:0:0:0:0:0:0:1.它是有效的环回地址形式.但它只是IPv6格式的localhost.localhost的IPv4地址是localhost的127.0.0.1IPv6地址0:0:0:0:0:0:0:1.

问题是URL https://localhost:8443/test/main.html默认匹配两个版本的IP协议.显然您的浏览器选择使用IPv6.

对于本地测试,请尝试使用文字地址127.0.0.1而不是名称localhost.或者,您可以在DNS设置中仅使用IPv4地址.