我有一个简单的Node.js程序在我的机器上运行,我想获得我的程序正在运行的PC的本地IP地址。我如何在Node.js中获得它?


当前回答

调用ifconfig是非常依赖于平台的,而且网络层知道套接字所在的IP地址,所以最好是询问它。

Node.js并没有公开这样做的直接方法,但是你可以打开任何套接字,并询问正在使用的本地IP地址。例如,打开www.google.com的套接字:

var net = require('net');
function getNetworkIP(callback) {
  var socket = net.createConnection(80, 'www.google.com');
  socket.on('connect', function() {
    callback(undefined, socket.address().address);
    socket.end();
  });
  socket.on('error', function(e) {
    callback(e, 'error');
  });
}

使用情况:

getNetworkIP(function (error, ip) {
    console.log(ip);
    if (error) {
        console.log('error:', error);
    }
});

其他回答

类似于其他答案,但更简洁:

'use strict';

const interfaces = require('os').networkInterfaces();

const addresses = Object.keys(interfaces)
  .reduce((results, name) => results.concat(interfaces[name]), [])
  .filter((iface) => iface.family === 'IPv4' && !iface.internal)
  .map((iface) => iface.address);

使用内部ip:

const internalIp = require("internal-ip")

console.log(internalIp.v4.sync())

下面是我获取本地IP地址的实用方法,假设您正在寻找一个IPv4地址,而机器只有一个真实的网络接口。可以很容易地对其进行重构,以返回多接口机器的IP地址数组。

function getIPAddress() {
  var interfaces = require('os').networkInterfaces();
  for (var devName in interfaces) {
    var iface = interfaces[devName];

    for (var i = 0; i < iface.length; i++) {
      var alias = iface[i];
      if (alias.family === 'IPv4' && alias.address !== '127.0.0.1' && !alias.internal)
        return alias.address;
    }
  }
  return '0.0.0.0';
}

这些信息可以在os.networkInterfaces()中找到,这是一个对象,它将网络接口名称映射到它的属性(例如,一个接口可以有几个地址):

'use strict';

const { networkInterfaces } = require('os');

const nets = networkInterfaces();
const results = Object.create(null); // Or just '{}', an empty object

for (const name of Object.keys(nets)) {
    for (const net of nets[name]) {
        // Skip over non-IPv4 and internal (i.e. 127.0.0.1) addresses
        // 'IPv4' is in Node <= 17, from 18 it's a number 4 or 6
        const familyV4Value = typeof net.family === 'string' ? 'IPv4' : 4
        if (net.family === familyV4Value && !net.internal) {
            if (!results[name]) {
                results[name] = [];
            }
            results[name].push(net.address);
        }
    }
}
// 'results'
{
  "en0": [
    "192.168.1.101"
  ],
  "eth0": [
    "10.0.0.101"
  ],
  "<network name>": [
    "<ip>",
    "<ip alias>",
    "<ip alias>",
    ...
  ]
}
// results["en0"][0]
"192.168.1.101"

公认的答案是异步的。我想要一个同步版本:

var os = require('os');
var ifaces = os.networkInterfaces();

console.log(JSON.stringify(ifaces, null, 4));

for (var iface in ifaces) {
  var iface = ifaces[iface];
  for (var alias in iface) {
    var alias = iface[alias];

    console.log(JSON.stringify(alias, null, 4));

    if ('IPv4' !== alias.family || alias.internal !== false) {
      debug("skip over internal (i.e. 127.0.0.1) and non-IPv4 addresses");
      continue;
    }
    console.log("Found IP address: " + alias.address);
    return alias.address;
  }
}
return false;