我是新android开发，我做它发送一个Android设备的IP地址，另外一个通过手机短信的应用程序。我需要得到知识产权的十进制这样192.168.0.4不是，我从下面的code有十六进制。任何想法如何做到这一点，感谢您的帮助。

 公共字符串getLocalIpAddress（）
    {
        尝试 {
            对于（枚举＆LT;的NetworkInterface＆GT; EN = NetworkInterface.getNetworkInterfaces（）; en.hasMoreElements（））{
                NetworkInterface的INTF = en.nextElement（）;
                对于（枚举＆LT; InetAddress类＆GT; enumIpAddr = intf.getInetAddresses（）; enumIpAddr.hasMoreElements（））{
                    InetAddress类InetAddress类= enumIpAddr.nextElement（）;
                    如果（！inetAddress.isLoopbackAddress（））{
                        返回inetAddress.getHostAddress（）的toString（）。

                    }
                 }
             }
         }赶上（SocketException前）{
             Log.e（TAG，ex.toString（））;
         }

         返回null;
    }
 

解决方案

 公共静态字符串getLocalIpv4Address（）{
    尝试 {
        字符串的IPv4;
        名单＆LT;的NetworkInterface＆GT; nilist = Collections.list（NetworkInterface.getNetworkInterfaces（））;
        如果（nilist.size（）大于0）{
            对于（的NetworkInterface妮：nilist）{
                名单＆LT; InetAddress类＆GT; ialist = Collections.list（ni.getInetAddresses（））;
                如果（ialist.size（）大于0）{
                    对于（InetAddress类地址：ialist）{
                        如果（address.isLoopbackAddress（）及！＆安培; InetAddressUtils.isIPv4Address（IPv4的= address.getHostAddress（）））{
                            返回IPv4的;
                        }
                    }
                }

            }
        }

    }赶上（SocketException前）{

    }
    返回 ;
}
 

如果这行吗？这个函数将返回的IPv4（以xxx.xxx.xxx.xxx模式），只有可用。

请注意，您提到的十六进制值应该是一个IPv6地址。

I am new to android development and I am doing an application which sends the IP address of an android device to another one by sms. I need to get the IP in decimal like this 192.168.0.4 not in hexadecimal which I got from the below code. any idea how to do that and thanks for the help.

    public String getLocalIpAddress()
    {
        try {
            for (Enumeration<NetworkInterface> en = NetworkInterface.getNetworkInterfaces(); en.hasMoreElements();) {
                NetworkInterface intf = en.nextElement();
                for (Enumeration<InetAddress> enumIpAddr = intf.getInetAddresses(); enumIpAddr.hasMoreElements();) {
                    InetAddress inetAddress = enumIpAddr.nextElement();
                    if (!inetAddress.isLoopbackAddress()) {
                        return inetAddress.getHostAddress().toString();

                    }
                 }
             }
         } catch (SocketException ex) {
             Log.e(TAG, ex.toString());
         }

         return null;
    } 

public static String getLocalIpv4Address(){
    try {
        String ipv4;
        List<NetworkInterface>  nilist = Collections.list(NetworkInterface.getNetworkInterfaces());
        if(nilist.size() > 0){
            for (NetworkInterface ni: nilist){
                List<InetAddress>  ialist = Collections.list(ni.getInetAddresses());
                if(ialist.size()>0){
                    for (InetAddress address: ialist){
                        if (!address.isLoopbackAddress() && InetAddressUtils.isIPv4Address(ipv4=address.getHostAddress())){ 
                            return ipv4;
                        }
                    }
                }

            }
        }

    } catch (SocketException ex) {

    }
    return "";
}

Should this be ok ? This function will return ipv4 (in xxx.xxx.xxx.xxx pattern) only if available.

Please note that those hexadecimal value you mentioned should be an ipv6 address.