[javascript] Extract hostname name from string

I would like to match just the root of a URL and not the whole URL from a text string. Given:

http://www.youtube.com/watch?v=ClkQA2Lb_iE
http://youtu.be/ClkQA2Lb_iE
http://www.example.com/12xy45
http://example.com/random

I want to get the 2 last instances resolving to the www.example.com or example.com domain.

I heard regex is slow and this would be my second regex expression on the page so If there is anyway to do it without regex let me know.

I'm seeking a JS/jQuery version of this solution.

function hostname(url) {
    var match = url.match(/:\/\/(www[0-9]?\.)?(.[^/:]+)/i);
    if ( match != null && match.length > 2 && typeof match[2] === 'string' && match[2].length > 0 ) return match[2];
}

The above code will successfully parse the hostnames for the following example urls:

http://WWW.first.com/folder/page.html first.com

http://mail.google.com/folder/page.html mail.google.com

https://mail.google.com/folder/page.html mail.google.com

http://www2.somewhere.com/folder/page.html?q=1 somewhere.com

https://www.another.eu/folder/page.html?q=1 another.eu

Original credit goes to: http://www.primaryobjects.com/CMS/Article145

in short way you can do like this

var url = "http://www.someurl.com/support/feature"

function getDomain(url){
  domain=url.split("//")[1];
  return domain.split("/")[0];
}
eg:
  getDomain("http://www.example.com/page/1")

  output:
   "www.example.com"

Use above function to get domain name

Try below code for exact domain name using regex,

String line = "http://www.youtube.com/watch?v=ClkQA2Lb_iE";

  String pattern3="([\\w\\W]\\.)+(.*)?(\\.[\\w]+)";

  Pattern r = Pattern.compile(pattern3);


  Matcher m = r.matcher(line);
  if (m.find( )) {

    System.out.println("Found value: " + m.group(2) );
  } else {
     System.out.println("NO MATCH");
  }

Code:

var regex = /\w+.(com|co\.kr|be)/ig;
var urls = ['http://www.youtube.com/watch?v=ClkQA2Lb_iE',
            'http://youtu.be/ClkQA2Lb_iE',
            'http://www.example.com/12xy45',
            'http://example.com/random'];


$.each(urls, function(index, url) {
    var convertedUrl = url.match(regex);
    console.log(convertedUrl);
});

Result:

youtube.com
youtu.be
example.com
example.com

2020 answer

You don't need any extra dependencies for this! Depending on whether you need to optimize for performance or not, there are two good solutions:

1. Use URL.hostname for readability

In the Babel era, the cleanest and easiest solution is to use URL.hostname.

const getHostname = (url) => {
  // use URL constructor and return hostname
  return new URL(url).hostname;
}

// tests
console.log(getHostname("https://stackoverflow.com/questions/8498592/extract-hostname-name-from-string/"));
console.log(getHostname("https://developer.mozilla.org/en-US/docs/Web/API/URL/hostname"));

URL.hostname is part of the URL API, supported by all major browsers except IE (caniuse). Use a URL polyfill if you need to support legacy browsers.

Using this solution will also give you access to other URL properties and methods. This will be useful if you also want to extract the URL's pathname or query string params, for example.

2. Use RegEx for performance

URL.hostname is faster than using the anchor solution or parseUri. However it's still much slower than gilly3's regex:

const getHostnameFromRegex = (url) => {
  // run against regex
  const matches = url.match(/^https?\:\/\/([^\/?#]+)(?:[\/?#]|$)/i);
  // extract hostname (will be null if no match is found)
  return matches && matches[1];
}

// tests
console.log(getHostnameFromRegex("https://stackoverflow.com/questions/8498592/extract-hostname-name-from-string/"));
console.log(getHostnameFromRegex("https://developer.mozilla.org/en-US/docs/Web/API/URL/hostname"));

Test it yourself on this jsPerf

TL;DR

If you need to process a very large number of URLs (where performance would be a factor), use RegEx. Otherwise, use URL.hostname.

A neat trick without using regular expressions:

var tmp        = document.createElement ('a');
;   tmp.href   = "http://www.example.com/12xy45";

// tmp.hostname will now contain 'www.example.com'
// tmp.host will now contain hostname and port 'www.example.com:80'

Wrap the above in a function such as the below and you have yourself a superb way of snatching the domain part out of an URI.

function url_domain(data) {
  var    a      = document.createElement('a');
         a.href = data;
  return a.hostname;
}

Okay, I know this is an old question, but I made a super-efficient url parser so I thought I'd share it.

As you can see, the structure of the function is very odd, but it's for efficiency. No prototype functions are used, the string doesn't get iterated more than once, and no character is processed more than necessary.

function getDomain(url) {
    var dom = "", v, step = 0;
    for(var i=0,l=url.length; i<l; i++) {
        v = url[i]; if(step == 0) {
            //First, skip 0 to 5 characters ending in ':' (ex: 'https://')
            if(i > 5) { i=-1; step=1; } else if(v == ':') { i+=2; step=1; }
        } else if(step == 1) {
            //Skip 0 or 4 characters 'www.'
            //(Note: Doesn't work with www.com, but that domain isn't claimed anyway.)
            if(v == 'w' && url[i+1] == 'w' && url[i+2] == 'w' && url[i+3] == '.') i+=4;
            dom+=url[i]; step=2;
        } else if(step == 2) {
            //Stop at subpages, queries, and hashes.
            if(v == '/' || v == '?' || v == '#') break; dom += v;
        }
    }
    return dom;
}

import URL from 'url';

const pathname = URL.parse(url).path;
console.log(url.replace(pathname, ''));

this takes care of both the protocol.

Here's the jQuery one-liner:

$('<a>').attr('href', url).prop('hostname');

Parsing a URL can be tricky because you can have port numbers and special chars. As such, I recommend using something like parseUri to do this for you. I doubt performance is going to be a issue unless you are parsing hundreds of URLs.

Well, doing using an regular expression will be a lot easier:

    mainUrl = "http://www.mywebsite.com/mypath/to/folder";
    urlParts = /^(?:\w+\:\/\/)?([^\/]+)(.*)$/.exec(mainUrl);
    host = Fragment[1]; // www.mywebsite.com

// use this if you know you have a subdomain
// www.domain.com -> domain.com
function getDomain() {
  return window.location.hostname.replace(/([a-zA-Z0-9]+.)/,"");
}

Try this:

var matches = url.match(/^https?\:\/\/([^\/?#]+)(?:[\/?#]|$)/i);
var domain = matches && matches[1];  // domain will be null if no match is found

If you want to exclude the port from your result, use this expression instead:

/^https?\:\/\/([^\/:?#]+)(?:[\/:?#]|$)/i

Edit: To prevent specific domains from matching, use a negative lookahead. (?!youtube.com)

/^https?\:\/\/(?!(?:www\.)?(?:youtube\.com|youtu\.be))([^\/:?#]+)(?:[\/:?#]|$)/i

I tried to use the Given solutions, the Chosen one was an overkill for my purpose and "Creating a element" one messes up for me.

It's not ready for Port in URL yet. I hope someone finds it useful

function parseURL(url){
    parsed_url = {}

    if ( url == null || url.length == 0 )
        return parsed_url;

    protocol_i = url.indexOf('://');
    parsed_url.protocol = url.substr(0,protocol_i);

    remaining_url = url.substr(protocol_i + 3, url.length);
    domain_i = remaining_url.indexOf('/');
    domain_i = domain_i == -1 ? remaining_url.length - 1 : domain_i;
    parsed_url.domain = remaining_url.substr(0, domain_i);
    parsed_url.path = domain_i == -1 || domain_i + 1 == remaining_url.length ? null : remaining_url.substr(domain_i + 1, remaining_url.length);

    domain_parts = parsed_url.domain.split('.');
    switch ( domain_parts.length ){
        case 2:
          parsed_url.subdomain = null;
          parsed_url.host = domain_parts[0];
          parsed_url.tld = domain_parts[1];
          break;
        case 3:
          parsed_url.subdomain = domain_parts[0];
          parsed_url.host = domain_parts[1];
          parsed_url.tld = domain_parts[2];
          break;
        case 4:
          parsed_url.subdomain = domain_parts[0];
          parsed_url.host = domain_parts[1];
          parsed_url.tld = domain_parts[2] + '.' + domain_parts[3];
          break;
    }

    parsed_url.parent_domain = parsed_url.host + '.' + parsed_url.tld;

    return parsed_url;
}

Running this:

parseURL('https://www.facebook.com/100003379429021_356001651189146');

Result:

Object {
    domain : "www.facebook.com",
    host : "facebook",
    path : "100003379429021_356001651189146",
    protocol : "https",
    subdomain : "www",
    tld : "com"
}

String.prototype.trim = function(){return his.replace(/^\s+|\s+$/g,"");}
function getHost(url){
    if("undefined"==typeof(url)||null==url) return "";
    url = url.trim(); if(""==url) return "";
    var _host,_arr;
    if(-1<url.indexOf("://")){
        _arr = url.split('://');
        if(-1<_arr[0].indexOf("/")||-1<_arr[0].indexOf(".")||-1<_arr[0].indexOf("\?")||-1<_arr[0].indexOf("\&")){
            _arr[0] = _arr[0].trim();
            if(0==_arr[0].indexOf("//")) _host = _arr[0].split("//")[1].split("/")[0].trim().split("\?")[0].split("\&")[0];
            else return "";
        }
        else{
            _arr[1] = _arr[1].trim();
            _host = _arr[1].split("/")[0].trim().split("\?")[0].split("\&")[0];
        }
    }
    else{
        if(0==url.indexOf("//")) _host = url.split("//")[1].split("/")[0].trim().split("\?")[0].split("\&")[0];
        else return "";
    }
    return _host;
}
function getHostname(url){
    if("undefined"==typeof(url)||null==url) return "";
    url = url.trim(); if(""==url) return "";
    return getHost(url).split(':')[0];
}
function getDomain(url){
    if("undefined"==typeof(url)||null==url) return "";
    url = url.trim(); if(""==url) return "";
    return getHostname(url).replace(/([a-zA-Z0-9]+.)/,"");
}

parse-domain - a very solid lightweight library

npm install parse-domain

const { fromUrl, parseDomain } = require("parse-domain");

Example 1

parseDomain(fromUrl("http://www.example.com/12xy45"))

{ type: 'LISTED',
  hostname: 'www.example.com',
  labels: [ 'www', 'example', 'com' ],
  icann:
   { subDomains: [ 'www' ],
     domain: 'example',
     topLevelDomains: [ 'com' ] },
  subDomains: [ 'www' ],
  domain: 'example',
  topLevelDomains: [ 'com' ] }

Example 2

parseDomain(fromUrl("http://subsub.sub.test.ExAmPlE.coM/12xy45"))

{ type: 'LISTED',
  hostname: 'subsub.sub.test.example.com',
  labels: [ 'subsub', 'sub', 'test', 'example', 'com' ],
  icann:
   { subDomains: [ 'subsub', 'sub', 'test' ],
     domain: 'example',
     topLevelDomains: [ 'com' ] },
  subDomains: [ 'subsub', 'sub', 'test' ],
  domain: 'example',
  topLevelDomains: [ 'com' ] }

Why?

Depending on the use case and volume I strongly recommend against solving this problem yourself using regex or other string manipulation means. The core of this problem is that you need to know all the gtld and cctld suffixes to properly parse url strings into domain and subdomains, these suffixes are regularly updated. This is a solved problem and not one you want to solve yourself (unless you are google or something). Unless you need the hostname or domain name in a pinch don't try and parse your way out of this one.

oneline with jquery

$('<a>').attr('href', document.location.href).prop('hostname');

This is not a full answer, but the below code should help you:

function myFunction() {
    var str = "https://www.123rf.com/photo_10965738_lots-oop.html";
    matches = str.split('/');
    return matches[2];
}

I would like some one to create code faster than mine. It help to improve my-self also.

If you end up on this page and you are looking for the best REGEX of URLS try this one:

^(?:https?:)?(?:\/\/)?([^\/\?]+)

https://regex101.com/r/pX5dL9/1

It works for urls without http:// , with http, with https, with just // and dont grab the path and query path as well.

Good Luck

Parse-Urls appears to be the JavaScript library with the most robust patterns

Here is a rundown of the features:

Chapter 1. Normalize or parse one URL

Chapter 2. Extract all URLs

Chapter 3. Extract URIs with certain names

Chapter 4. Extract all fuzzy URLs

Chapter 5. Highlight all URLs in texts

Chapter 6. Extract all URLs in raw HTML or XML

There is no need to parse the string, just pass your URL as an argument to URL constructor:

const url = 'http://www.youtube.com/watch?v=ClkQA2Lb_iE';
const { hostname } = new URL(url);

console.assert(hostname === 'www.youtube.com');

All url properties, no dependencies, no JQuery, easy to understand

This solution gives your answer plus additional properties. No JQuery or other dependencies required, paste and go.

Usage

getUrlParts("https://news.google.com/news/headlines/technology.html?ned=us&hl=en")

Output

{
  "origin": "https://news.google.com",
  "domain": "news.google.com",
  "subdomain": "news",
  "domainroot": "google.com",
  "domainpath": "news.google.com/news/headlines",
  "tld": ".com",
  "path": "news/headlines/technology.html",
  "query": "ned=us&hl=en",
  "protocol": "https",
  "port": 443,
  "parts": [
    "news",
    "google",
    "com"
  ],
  "segments": [
    "news",
    "headlines",
    "technology.html"
  ],
  "params": [
    {
      "key": "ned",
      "val": "us"
    },
    {
      "key": "hl",
      "val": "en"
    }
  ]
}

Code
The code is designed to be easy to understand rather than super fast. It can be called easily 100 times per second, so it's great for front end or a few server usages, but not for high volume throughput.

function getUrlParts(fullyQualifiedUrl) {
    var url = {},
        tempProtocol
    var a = document.createElement('a')
    // if doesn't start with something like https:// it's not a url, but try to work around that
    if (fullyQualifiedUrl.indexOf('://') == -1) {
        tempProtocol = 'https://'
        a.href = tempProtocol + fullyQualifiedUrl
    } else
        a.href = fullyQualifiedUrl
    var parts = a.hostname.split('.')
    url.origin = tempProtocol ? "" : a.origin
    url.domain = a.hostname
    url.subdomain = parts[0]
    url.domainroot = ''
    url.domainpath = ''
    url.tld = '.' + parts[parts.length - 1]
    url.path = a.pathname.substring(1)
    url.query = a.search.substr(1)
    url.protocol = tempProtocol ? "" : a.protocol.substr(0, a.protocol.length - 1)
    url.port = tempProtocol ? "" : a.port ? a.port : a.protocol === 'http:' ? 80 : a.protocol === 'https:' ? 443 : a.port
    url.parts = parts
    url.segments = a.pathname === '/' ? [] : a.pathname.split('/').slice(1)
    url.params = url.query === '' ? [] : url.query.split('&')
    for (var j = 0; j < url.params.length; j++) {
        var param = url.params[j];
        var keyval = param.split('=')
        url.params[j] = {
            'key': keyval[0],
            'val': keyval[1]
        }
    }
    // domainroot
    if (parts.length > 2) {
        url.domainroot = parts[parts.length - 2] + '.' + parts[parts.length - 1];
        // check for country code top level domain
        if (parts[parts.length - 1].length == 2 && parts[parts.length - 1].length == 2)
            url.domainroot = parts[parts.length - 3] + '.' + url.domainroot;
    }
    // domainpath (domain+path without filenames) 
    if (url.segments.length > 0) {
        var lastSegment = url.segments[url.segments.length - 1]
        var endsWithFile = lastSegment.indexOf('.') != -1
        if (endsWithFile) {
            var fileSegment = url.path.indexOf(lastSegment)
            var pathNoFile = url.path.substr(0, fileSegment - 1)
            url.domainpath = url.domain
            if (pathNoFile)
                url.domainpath = url.domainpath + '/' + pathNoFile
        } else
            url.domainpath = url.domain + '/' + url.path
    } else
        url.domainpath = url.domain
    return url
}

Was looking for a solution to this problem today. None of the above answers seemed to satisfy. I wanted a solution that could be a one liner, no conditional logic and nothing that had to be wrapped in a function.

Here's what I came up with, seems to work really well:

hostname="http://www.example.com:1234"
hostname.split("//").slice(-1)[0].split(":")[0].split('.').slice(-2).join('.')   // gives "example.com"

May look complicated at first glance, but it works pretty simply; the key is using 'slice(-n)' in a couple of places where the good part has to be pulled from the end of the split array (and [0] to get from the front of the split array).

Each of these tests return "example.com":

"http://example.com".split("//").slice(-1)[0].split(":")[0].split('.').slice(-2).join('.')
"http://example.com:1234".split("//").slice(-1)[0].split(":")[0].split('.').slice(-2).join('.')
"http://www.example.com:1234".split("//").slice(-1)[0].split(":")[0].split('.').slice(-2).join('.')
"http://foo.www.example.com:1234".split("//").slice(-1)[0].split(":")[0].split('.').slice(-2).join('.')

My code looks like this. Regular expressions can come in many forms, and here are my test cases I think it's more scalable.

function extractUrlInfo(url){_x000D_
  let reg = /^((?<protocol>http[s]?):\/\/)?(?<host>((\d{1,2}|1\d\d|2[0-4]\d|25[0-5])\.(\d{1,2}|1\d\d|2[0-4]\d|25[0-5])\.(\d{1,2}|1\d\d|2[0-4]\d|25[0-5])\.(\d{1,2}|1\d\d|2[0-4]\d|25[0-5])|[-a-zA-Z0-9@:%._\+~#=]{1,256}\.[a-zA-Z0-9()]{1,6}\b([-a-zA-Z0-9()@:%_\+.~#?&//=]*)))(\:(?<port>[0-9]|[1-9]\d|[1-9]\d{2}|[1-9]\d{3}|[1-5]\d{4}|6[0-4]\d{3}|65[0-4]\d{2}|655[0-2]\d|6553[0-5]))?$/_x000D_
  return reg.exec(url).groups_x000D_
}_x000D_
_x000D_
var url = "https://192.168.1.1:1234"_x000D_
console.log(extractUrlInfo(url))_x000D_
var url = "https://stackoverflow.com/questions/8498592/extract-hostname-name-from-string"_x000D_
console.log(extractUrlInfo(url))

I personally researched a lot for this solution, and the best one I could find is actually from CloudFlare's "browser check":

function getHostname(){  
            secretDiv = document.createElement('div');
            secretDiv.innerHTML = "<a href='/'>x</a>";
            secretDiv = secretDiv.firstChild.href;
            var HasHTTPS = secretDiv.match(/https?:\/\//)[0];
            secretDiv = secretDiv.substr(HasHTTPS.length);
            secretDiv = secretDiv.substr(0, secretDiv.length - 1);
            return(secretDiv);  
}  

getHostname();

I rewritten variables so it is more "human" readable, but it does the job better than expected.

Just use the URL() constructor:

new URL(url).host