umair umair - 1 year ago 179
Javascript Question

how to edit local ip Address

By using the code below i am able to get local IP.but i have to edit this IP address so that it appears in pattern like this( x.x.x.200 ).
1st three octet should be same as per router but last one should be constant(200).
Your responses will be highly appreciated.

var findIP = new Promise(r => {
var w = window,
a = new(w.RTCPeerConnection || w.mozRTCPeerConnection || w.webkitRTCPeerConnection)({ iceServers: [] }),
b = () => {};
a.createOffer(c => a.setLocalDescription(c, b, b), b);
a.onicecandidate = c => {
try { c.candidate.candidate.match(/([0-9]{1,3}(\.[0-9]{1,3}){3}|[a-f0-9]{1,4}(:[a-f0-9]{1,4}){7})/g).forEach(r) } catch (e) {}

/*Usage example*/
findIP.then(ip => document.write('your ip: ', ip)).catch(e => console.error(e))

How to get client's IP address using javascript only?

Answer Source

There are many ways to achieve this. its better if you only get first 3 octate during your initial function itself.

But following will also do the job:

var findIP = new Promise(r=>{var w=window,a=new (w.RTCPeerConnection||w.mozRTCPeerConnection||w.webkitRTCPeerConnection)({iceServers:[]}),b=()=>{};a.createDataChannel("");a.createOffer(c=>a.setLocalDescription(c,b,b),b);a.onicecandidate=c=>{try{c.candidate.candidate.match(/([0-9]{1,3}(\.[0-9]{1,3}){3}|[a-f0-9]{1,4}(:[a-f0-9]{1,4}){7})/g).forEach(r)}catch(e){}}})

/*Usage example*/
findIP.then(ip => console.log('your ip: ', ip.split('.')[0]+'.'+ip.split('.')[1]+'.'+ip.split('.')[2]+'.200')).catch(e => console.error(e))

EDIT: You can change actual code using webrtc to give the ip u needed as follows:

var findIP = new Promise(r=>{var w=window,a=new (w.RTCPeerConnection||w.mozRTCPeerConnection||w.webkitRTCPeerConnection)({iceServers:[]}),b=()=>{};a.createDataChannel("");a.createOffer(c=>a.setLocalDescription(c,b,b),b);a.onicecandidate=c=>{try{c.candidate.candidate.match(/\b(\d{1,3}\.){2}\d{1,3}\b/).forEach(r)}catch(e){}}})

findIP.then(ip => document.write('your ip: ', ip+'.200')).catch(e => console.error(e))
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