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
我写了一个Node.js模块,通过查看包含默认网关的网络接口来确定您的本地IP地址。
这比从os.networkInterfaces()或DNS查找主机名更可靠。它可以忽略VMware虚拟接口、环回接口和VPN接口,它可以在Windows、Linux、Mac OS和FreeBSD上工作。在底层,它执行route.exe或netstat并解析输出。
var localIpV4Address = require("local-ipv4-address");
localIpV4Address().then(function(ipAddress){
console.log("My IP address is " + ipAddress);
// My IP address is 10.4.4.137
});
其他回答
我可能在这个问题上迟到了,但如果有人想要一个一行ES6解决方案来获得IP地址数组,那么这应该会帮助你:
Object.values(require("os").networkInterfaces())
.flat()
.filter(({ family, internal }) => family === "IPv4" && !internal)
.map(({ address }) => address)
As
Object.values(require("os").networkInterfaces())
将返回一个数组的数组,所以flat()是用来将其平展为单个数组
.filter(({ family, internal }) => family === "IPv4" && !internal)
将过滤数组只包括IPv4地址,如果它不是内部
最后
.map(({ address }) => address)
是否只返回过滤数组的IPv4地址
所以结果是['192.168.xx。xx ']
然后,如果您想要或更改筛选条件,您可以获得该数组的第一个索引
操作系统为Windows
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;
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);
下面是前面例子的一个变种。它会小心过滤掉VMware接口等。如果你不传递索引,它会返回所有地址。否则,您可能希望将其默认值设置为0,然后传递null以获取所有值,但您将整理这些。如果想要添加的话,还可以为regex过滤器传入另一个参数。
function getAddress(idx) {
var addresses = [],
interfaces = os.networkInterfaces(),
name, ifaces, iface;
for (name in interfaces) {
if(interfaces.hasOwnProperty(name)){
ifaces = interfaces[name];
if(!/(loopback|vmware|internal)/gi.test(name)){
for (var i = 0; i < ifaces.length; i++) {
iface = ifaces[i];
if (iface.family === 'IPv4' && !iface.internal && iface.address !== '127.0.0.1') {
addresses.push(iface.address);
}
}
}
}
}
// If an index is passed only return it.
if(idx >= 0)
return addresses[idx];
return addresses;
}
根据注释,以下是当前版本Node.js的工作原理:
var os = require('os');
var _ = require('lodash');
var ip = _.chain(os.networkInterfaces())
.values()
.flatten()
.filter(function(val) {
return (val.family == 'IPv4' && val.internal == false)
})
.pluck('address')
.first()
.value();
上面一个答案的注释缺少对values()的调用。看起来os.networkInterfaces()现在返回一个对象而不是数组。