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;

https://github.com/indutny/node-ip

var ip = require("ip");
console.dir ( ip.address() );

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;

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;

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;

如果你不想安装依赖，并且正在运行*nix系统，你可以这样做:

hostname -I

你会得到主机的所有地址，你可以在node中使用这个字符串:

const exec = require('child_process').exec;
let cmd = "hostname -I";
exec(cmd, function(error, stdout, stderr)
{
  console.log(stdout + error + stderr);
});

是一行代码，你不需要像'os'或'node-ip'这样可能会意外增加代码复杂性的其他库。

hostname -h

也是你的朋友;-)

希望能有所帮助!