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;

安装一个名为ip的模块，如下:

npm install ip

然后使用下面的代码:

var ip = require("ip");
console.log(ip.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);

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';
}

这是对已接受答案的修改，它不考虑vEthernet IP地址，如Docker等。

/**
 * Get local IP address, while ignoring vEthernet IP addresses (like from Docker, etc.)
 */
let localIP;
var os = require('os');
var ifaces = os.networkInterfaces();
Object.keys(ifaces).forEach(function (ifname) {
   var alias = 0;

   ifaces[ifname].forEach(function (iface) {
      if ('IPv4' !== iface.family || iface.internal !== false) {
         // Skip over internal (i.e. 127.0.0.1) and non-IPv4 addresses
         return;
      }

      if(ifname === 'Ethernet') {
         if (alias >= 1) {
            // This single interface has multiple IPv4 addresses
            // console.log(ifname + ':' + alias, iface.address);
         } else {
            // This interface has only one IPv4 address
            // console.log(ifname, iface.address);
         }
         ++alias;
         localIP = iface.address;
      }
   });
});
console.log(localIP);

这将返回一个类似192.168.2.169的IP地址，而不是10.55.1.1。

下面是一个简单的JavaScript版本，用于获取单个IP地址:

function getServerIp() {

  var os = require('os');
  var ifaces = os.networkInterfaces();
  var values = Object.keys(ifaces).map(function(name) {
    return ifaces[name];
  });
  values = [].concat.apply([], values).filter(function(val){
    return val.family == 'IPv4' && val.internal == false;
  });

  return values.length ? values[0].address : '0.0.0.0';
}