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';
}

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';
}

谷歌在搜索“Node.js获取服务器IP”时引导我到这个问题，所以让我们为那些试图在他们的Node.js服务器程序中实现这一点的人提供一个替代答案(可能是原始海报的情况)。

在最简单的情况下，服务器只绑定到一个IP地址，应该不需要确定IP地址，因为我们已经知道将它绑定到哪个地址(例如，传递给listen()函数的第二个参数)。

在不太简单的情况下，服务器绑定到多个IP地址，我们可能需要确定客户端连接到的接口的IP地址。正如Tor Valamo所简单建议的，现在，我们可以很容易地从连接的套接字及其localAddress属性中获得这些信息。

例如，如果程序是web服务器:

var http = require("http")

http.createServer(function (req, res) {
    console.log(req.socket.localAddress)
    res.end(req.socket.localAddress)
}).listen(8000)

如果它是一个通用TCP服务器:

var net = require("net")

net.createServer(function (socket) {
    console.log(socket.localAddress)
    socket.end(socket.localAddress)
}).listen(8000)

在运行服务器程序时，该解决方案提供了非常高的可移植性、准确性和效率。

详情请参见:

http://nodejs.org/api/net.html http://nodejs.org/api/http.html

很多时候，我发现有多个内部和外部面向接口可用(例如:10.0.75.1,172.100.0.1,192.168.2.3)，而我真正想要的是外部接口(172.100.0.1)。

如果其他人也有类似的担忧，这里还有一个关于这个问题的看法，希望能有所帮助……

const address = Object.keys(os.networkInterfaces())
    // flatten interfaces to an array
    .reduce((a, key) => [
        ...a,
        ...os.networkInterfaces()[key]
    ], [])
    // non-internal ipv4 addresses only
    .filter(iface => iface.family === 'IPv4' && !iface.internal)
    // project ipv4 address as a 32-bit number (n)
    .map(iface => ({...iface, n: (d => ((((((+d[0])*256)+(+d[1]))*256)+(+d[2]))*256)+(+d[3]))(iface.address.split('.'))}))
    // set a hi-bit on (n) for reserved addresses so they will sort to the bottom
    .map(iface => iface.address.startsWith('10.') || iface.address.startsWith('192.') ? {...iface, n: Math.pow(2,32) + iface.n} : iface)
    // sort ascending on (n)
    .sort((a, b) => a.n - b.n)
    [0]||{}.address;

这里有一个简洁的小命令行，它实现了这个功能:

const ni = require('os').networkInterfaces();
Object
  .keys(ni)
  .map(interf =>
    ni[interf].map(o => !o.internal && o.family === 'IPv4' && o.address))
  .reduce((a, b) => a.concat(b))
  .filter(o => o)
  [0];

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);