使用Java获取当前计算机的IP地址 [英] Getting the IP address of the current machine using Java

问题描述

我正在尝试开发一个系统，其中有不同的节点在不同的系统上或在同一系统上的不同端口上运行。

I am trying to develop a system where there are different nodes that are run on different system or on different ports on the same system.

现在所有节点都创建一个Socket，目标IP作为特殊节点的IP，称为自举节点。然后节点创建自己的 ServerSocket 并开始侦听连接。

Now all the nodes create a Socket with a target IP as the IP of a special node known as a bootstrapping node. The nodes then create their own ServerSocket and start listening for connections.

引导节点维护一个节点列表并在被查询时返回它们。

The bootstrapping node maintains a list of Nodes and returns them on being queried.

现在我需要的是节点必须将其IP注册到引导节点。一旦客户端连接到bootstrapping节点的 ServerSocket ，我就尝试使用 cli.getInetAddress（）但这不起作用。

Now what I need is the node must register its IP to the bootstrapping node. I tried using cli.getInetAddress() once the client connects to the ServerSocket of bootstrapping node but that didn't work.

1. 我需要客户注册其PPP IP（如果有）;


2. 否则LAN IP如果可供使用的话;


3. 否则它必须注册127.0.0.1，假设它是同一台计算机。

使用代码：

System.out.println(Inet4Address.getLocalHost().getHostAddress());

或

System.out.println(InetAddress.getLocalHost().getHostAddress());

我的PPP连接IP地址是：117.204.44.192，但上面的回报是192.168.1.2

My PPP Connection IP address is: 117.204.44.192 but the above returns me 192.168.1.2

编辑

我使用以下代码：

Enumeration e = NetworkInterface.getNetworkInterfaces();
while(e.hasMoreElements())
{
    NetworkInterface n = (NetworkInterface) e.nextElement();
    Enumeration ee = n.getInetAddresses();
    while (ee.hasMoreElements())
    {
        InetAddress i = (InetAddress) ee.nextElement();
        System.out.println(i.getHostAddress());
    }
}

我能够获得所有相关的IP地址 NetworkInterface s，但我如何区分它们？这是我得到的输出：

I am able to get all the IP addresses associated all NetworkInterfaces, but how do I distinguish them? This is the output I am getting:

127.0.0.1
192.168.1.2
192.168.56.1
117.204.44.19

推荐答案

import java.net.DatagramSocket;

try(final DatagramSocket socket = new DatagramSocket()){
  socket.connect(InetAddress.getByName("8.8.8.8"), 10002);
  ip = socket.getLocalAddress().getHostAddress();
}

这种方式适用于有多个网络接口的情况。它始终返回首选的出站IP。目标 8.8.8.8 不需要可访问。

This way works well when there are multiple network interfaces. It always returns the preferred outbound IP. The destination 8.8.8.8 is not needed to be reachable.

在UDP套接字上连接会产生以下影响：它设置发送/接收的目的地，丢弃所有数据包从其他地址， - 我们使用的 - 将套接字转换为连接状态，设置其适当的字段。这包括根据系统的路由表检查到目的地的路由是否存在，并相应地设置本地端点。最后一部分似乎没有正式的文档，但它看起来像是Berkeley套接字API（UDP连接状态的副作用）的一个完整特征，它可以在Windows和Linux中跨版本和发行版可靠地工作。

Connect on a UDP socket has the following effect: it sets the destination for Send/Recv, discards all packets from other addresses, and - which is what we use - transfers the socket into "connected" state, settings its appropriate fields. This includes checking the existence of the route to the destination according to the system's routing table and setting the local endpoint accordingly. The last part seems to be undocumented officially but it looks like an integral trait of Berkeley sockets API (a side effect of UDP "connected" state) that works reliably in both Windows and Linux across versions and distributions.

因此，此方法将提供用于连接指定远程主机的本地地址。没有建立真正的连接，因此指定的远程IP无法访问。

So, this method will give the local address that would be used to connect to the specified remote host. There is no real connection established, hence the specified remote ip can be unreachable.

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