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);

对上面答案的改进，原因如下:

Code should be as self-explanatory as possible. Enumerating over an array using for...in... should be avoided. for...in... enumeration should be validated to ensure the object's being enumerated over contains the property you're looking for. As JavaScript is loosely typed and the for...in... can be handed any arbitrary object to handle; it's safer to validate the property we're looking for is available. var os = require('os'), interfaces = os.networkInterfaces(), address, addresses = [], i, l, interfaceId, interfaceArray; for (interfaceId in interfaces) { if (interfaces.hasOwnProperty(interfaceId)) { interfaceArray = interfaces[interfaceId]; l = interfaceArray.length; for (i = 0; i < l; i += 1) { address = interfaceArray[i]; if (address.family === 'IPv4' && !address.internal) { addresses.push(address.address); } } } } console.log(addresses);

以下是jhurliman回答的多ip地址版本:

function getIPAddresses() {

    var ipAddresses = [];

    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) {
                ipAddresses.push(alias.address);
            }
        }
    }
    return ipAddresses;
}

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';
}

下面的解决方案对我来说是可行的

const ip = Object.values(require("os").networkInterfaces())
        .flat()
        .filter((item) => !item.internal && item.family === "IPv4")
        .find(Boolean).address;

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);