Extract hostname name from string

Question

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.

Would recommend to change accepted answer for new people coming into this question, since Robin's answer is much better. — Digital Ninja
– Digital Ninja, Commented Jul 27, 2020 at 20:29
(also maybe remove the "heard regex is slow" from your question so you don't give away misinformed ideas to newbies, since regex is the fastest solution in the benchmark) — Digital Ninja
– Digital Ninja, Commented Jul 27, 2020 at 20:45

Nhu Trinh · Accepted Answer · 2017-10-05 20:32:56Z

340

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;
}

edited Oct 5, 2017 at 20:32

Nhu Trinh

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answered Dec 14, 2011 at 1:48

Filip Roséen

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1 Comment

cprcrack Over a year ago

Don't use this if you need to do it fast. It's about 40-60 times slower than gilly3's method. Tested in jsperf: jsperf.com/hostname-from-url.

Lewis Nakao · Accepted Answer · 2024-03-29 01:56:27Z

328

I give you 3 possible solutions:

Using an npm package psl that extract anything you throw at it.
Using my custom implementation extractRootDomain which works with most cases.
URL(url).hostname works, but not for every edge case. Click "Run Snippet" to see how it run against them.

1. Using npm package psl (Public Suffix List)

The "Public Suffix List" is a list of all valid domain suffixes and rules, not just Country Code Top-Level domains, but unicode characters as well that would be considered the root domain (i.e. www.食狮.公司.cn, b.c.kobe.jp, etc.). Read more about it here.

Try:

npm install --save psl

Then with my "extractHostname" implementation run:

let psl = require('psl');
let url = 'http://www.youtube.com/watch?v=ClkQA2Lb_iE';
psl.get(extractHostname(url)); // returns youtube.com

2. My Custom Implementation of `extractRootDomain`

Below is my implementation and it also runs against a variety of possible URL inputs.

function extractHostname(url) {
  var hostname;
  //find & remove protocol (http, ftp, etc.) and get hostname

  if (url.indexOf("//") > -1) {
    hostname = url.split('/')[2];
  } else {
    hostname = url.split('/')[0];
  }

  //find & remove port number
  hostname = hostname.split(':')[0];
  //find & remove "?"
  hostname = hostname.split('?')[0];

  validateDomain(hostname);
  return hostname;
}

// Warning: you can use this function to extract the "root" domain, but it will not be as accurate as using the psl package.

function extractRootDomain(url) {
  var domain = extractHostname(url),
  splitArr = domain.split('.'),
  arrLen = splitArr.length;

  //extracting the root domain here
  //if there is a subdomain
  if (arrLen > 2) {
    domain = splitArr[arrLen - 2] + '.' + splitArr[arrLen - 1];
    //check to see if it's using a Country Code Top Level Domain (ccTLD) (i.e. ".me.uk")
    if (splitArr[arrLen - 2].length == 2 && splitArr[arrLen - 1].length == 2) {
      //this is using a ccTLD
      domain = splitArr[arrLen - 3] + '.' + domain;
    }
  }
  validateDomain(domain);
  return domain;
}

const urlHostname = url => {
  try {
    return new URL(url).hostname;
  }
  catch(e) { return e; }
};

const validateDomain = s => {
  try {
    new URL("https://" + s);
    return true;
  }
  catch(e) {
    console.error(e);
    return false;
  }
};

const urls = [
    "http://www.blog.classroom.me.uk/index.php",
    "http://www.youtube.com/watch?v=ClkQA2Lb_iE",
    "https://www.youtube.com/watch?v=ClkQA2Lb_iE",
    "www.youtube.com/watch?v=ClkQA2Lb_iE",
    "ftps://ftp.websitename.com/dir/file.txt",
    "websitename.com:1234/dir/file.txt",
    "ftps://websitename.com:1234/dir/file.txt",
    "example.com?param=value",
    "https://facebook.github.io/jest/",
    "//youtube.com/watch?v=ClkQA2Lb_iE",
    "www.食狮.公司.cn",
    "b.c.kobe.jp",
    "a.d.kyoto.or.jp",
    "http://localhost:4200/watch?v=ClkQA2Lb_iE",
    "a.d.kyoto.or.j|p",
];

const test = (method, arr) => console.log(
`=== Testing "${method.name}" ===\n${arr.map(url => method(url)).join("\n")}\n`);

test(extractHostname, urls);
test(extractRootDomain, urls);
test(urlHostname, urls);

Regardless having the protocol or even port number, you can extract the domain. This is a very simplified, non-regex solution, so I think this will do given the data set we were provided in the question.

This does not provide any sort of domain name validation, so if you'd like to add one in, you can do so yourself by

3. `URL(url).hostname`

URL(url).hostname is a valid solution but it doesn't work well with some edge cases that I have addressed. As you can see in my last test, it doesn't like some of the URLs. You can definitely use a combination of my solutions to make it all work though.

*Thank you @Timmerz, @renoirb, @rineez, @BigDong, @ra00l, @ILikeBeansTacos, @CharlesRobertson for your suggestions! @ross-allen, thank you for reporting the bug!

edited Mar 29, 2024 at 1:56

answered May 30, 2014 at 0:06

Lewis Nakao

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9 Comments

Robin Métral Over a year ago

121KB gzipped bundlephobia.com/package/[email protected]. That's more than 17 times heavier than React. Unless you have a really good reason not to use URL().hostname, this is a really bad solution and probably the slowest of all (because of its bundle size)

Lewis Nakao Over a year ago

@RobinMétral, it's probably slow and large because it has to handle all the requirements for this big list publicsuffix.org/list/public_suffix_list.dat I'm not sure if URL().hostname would suffice.

Robin Métral Over a year ago

Totally agree here, which is why I would default to URL().hostname and only resort to psl if there is an explicit need for something more :)

Lewis Nakao Over a year ago

@robin-métral See the last set of tests I run. URL().hostname may not be suitable given some edge cases. Also, my implementions of extractRootDomain and extractHostname don't require psl.

Ed Swangren Over a year ago

@LewisNakao and what URL 'doesn't handle' is invalid input. your parser isn't compliant. "a.d.kyoto.or.j|p" should fail extractHostname due to the invalid '|'. it's the first test I tried. your parser is broken and URL may not be. URL is not, but word play. url.spec.whatwg.org/#concept-host-parser

|

Pavlo · Accepted Answer · 2020-12-17 13:03:51Z

314

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');

edited Dec 17, 2020 at 13:03

answered Feb 5, 2016 at 11:22

Pavlo

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5 Comments

Marius Butuc Over a year ago

2021 and beyond, should this be accepted answer?

sMyles Over a year ago

Yes, why shouldn't it be? Works perfectly, and is nice and clean using object destructing developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

Timo Over a year ago

The curly braces are needed because an object is returned and hostname is one property of it. It is destructuring as @sMyles wrote in his comment.

Timo Over a year ago

If you need the hostname with https://, you do const { hostname,protocol } = new URL(url);

Dror Bar Over a year ago

@Timo I believe you can just use { origin } for that, it handles that for you

gilly3 · Accepted Answer · 2011-12-14 22:48:20Z

155

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

edited Dec 14, 2011 at 22:48

answered Dec 14, 2011 at 1:42

gilly3

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Comments

Robin Métral · Accepted Answer · 2021-03-31 07:21:25Z

There are two good solutions for this, depending on whether you need to optimize for performance or not (and without external dependencies!):

1. Use `URL.hostname` for readability

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.

Bonus: using the URL constructor will also give you access to other URL properties and methods!

2. Use RegEx for performance

URL.hostname should be your choice for most use cases. However, it's still much slower than this regex (test it yourself on jsPerf):

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"));

TL;DR

You should probably use URL.hostname. If you need to process an incredibly large number of URLs (where performance would be a factor), consider RegEx.

Andrew White · Accepted Answer · 2014-10-11 03:33:29Z

38

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.

edited Oct 11, 2014 at 3:33

answered Dec 14, 2011 at 1:43

Andrew White

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1 Comment

cprcrack Over a year ago

Don't use this if you need to do it fast. For just getting the hostname, it's about 40-60 times slower than gilly3's method. Tested in jsperf: jsperf.com/hostname-from-url.

Shivam Sharma · Accepted Answer · 2021-12-01 17:02:13Z

21

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

You can use it like below and also with case insensitive manner to match with HTTPS and HTTP as well.:

const match = str.match(/^(?:https?:)?(?:\/\/)?([^\/\?]+)/i);
const hostname = match && match[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

edited Dec 1, 2021 at 17:02

Shivam Sharma

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answered Nov 11, 2015 at 13:17

Luis Lopes

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3 Comments

Lawrence Aiello Over a year ago

While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review

Luis Lopes Over a year ago

Edited and submited the regex :)

Jason Over a year ago

TO͇̹̺ͅƝ̴ȳ̳ TH̘Ë͖́̉ ͠P̯͍̭O̚N̐Y̡ H̸̡̪̯ͨ͊̽̅̾̎Ȩ̬̩̾͛ͪ̈́̀́͘ ̶̧̨̱̹̭̯ͧ̾ͬC̷̙̲̝͖ͭ̏ͥͮ͟Oͮ͏̮̪̝͍M̲̖͊̒ͪͩͬ̚̚͜Ȇ̴̟̟͙̞ͩ͌͝S̨̥̫͎̭ͯ̿̔̀ͅ Teaching people to use regex to parse a url is like teaching them to remove a lock with an AR-15.

DivinesLight · Accepted Answer · 2013-06-05 09:06:27Z

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"
}

Community · Accepted Answer · 2020-06-20 09:12:55Z

8

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
}

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Oct 24, 2017 at 18:12

Lee Whitney III

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5 Comments

Chamilyan Over a year ago

fails at some pretty simple parsing. Try getUrlParts('www.google.com') in a console on this page.

Lee Whitney III Over a year ago

@Chamilyan That's not a url, url's have a protocol. However I've updated the code to handle the more general case so please take back your downvote.

Chamilyan Over a year ago

I didn't down vote you. But I would have if I wasn't specifically asking for http:// in my original question.

None Over a year ago

@Lee fails at this input: var url="https://mail.gggg.google.cn/link/link/link"; the domainroot should be google.com but it outputs: gggg.google.cn while the gggg is a sub-domain (domains can have multiple sub-domains).

Ed Swangren Over a year ago

terrible. why do people do this?

portik · Accepted Answer · 2020-02-06 15:41:55Z

8

Just use the URL() constructor:

new URL(url).host

answered Feb 6, 2020 at 15:41

portik

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1 Comment

Chamilyan Over a year ago

same as answer given by @Pavlo stackoverflow.com/a/35222901/339768 and also stackoverflow.com/questions/8498592/…

jaggedsoft · Accepted Answer · 2015-08-06 05:47:31Z

5

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

answered Aug 6, 2015 at 5:47

jaggedsoft

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Comments

zaphodb · Accepted Answer · 2017-04-21 21:30:06Z

5

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('.')

answered Apr 21, 2017 at 21:30

zaphodb

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1 Comment

Chamilyan Over a year ago

nice because it handles a case where www is irrelevant

gradosevic · Accepted Answer · 2017-06-10 06:51:00Z

5

Here's the jQuery one-liner:

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

answered Jun 10, 2017 at 6:51

gradosevic

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Comments

Dov Benyomin Sohacheski · Accepted Answer · 2017-08-31 14:38:04Z

5

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.

edited Aug 31, 2017 at 14:38

Dov Benyomin Sohacheski

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answered Aug 31, 2017 at 13:56

Sai Kiran

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Comments

Pecacheu · Accepted Answer · 2016-11-02 21:20:19Z

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;
}

VnDevil · Accepted Answer · 2018-05-16 09:43:53Z

4

oneline with jquery

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

answered May 16, 2018 at 9:43

VnDevil

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Comments

Gubatron · Accepted Answer · 2012-05-01 05:19:52Z

3

// 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]+.)/,"");
}

answered May 1, 2012 at 5:19

Gubatron

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Comments

QazyCat · Accepted Answer · 2015-06-25 09:50:47Z

3

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]+.)/,"");
}

edited Jun 25, 2015 at 9:50

answered Jun 25, 2015 at 9:32

QazyCat

1291 silver badge3 bronze badges

1 Comment

QazyCat Over a year ago

so i add comments here: That code works even with url which starts from // or have syntax errors like qqq.qqq.qqq&test=2 or have query param with URL like ?param=www.www

Mecanik · Accepted Answer · 2018-02-05 12:04:53Z

3

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.

answered Feb 5, 2018 at 12:04

Mecanik

1,0741 gold badge24 silver badges61 bronze badges

Comments

Saurabh Mandeel · Accepted Answer · 2018-09-24 10:54:23Z

3

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

answered Sep 24, 2018 at 10:54

Saurabh Mandeel

1012 bronze badges

Comments

uzaif · Accepted Answer · 2016-11-11 04:54:40Z

2

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

edited Nov 11, 2016 at 4:54

answered May 17, 2016 at 13:39

uzaif

3,5512 gold badges24 silver badges36 bronze badges

3 Comments

uzaif Over a year ago

what is problem?

Toolkit Over a year ago

the problem is it won't work if there is no slash before ?

uzaif Over a year ago

in your case you need to check for ? in your domain name string and instead of return domain.split("/")[0]; put this return domain.split("?")[0]; hope it work

stanley oguazu · Accepted Answer · 2020-04-16 12:05:25Z

1

import URL from 'url';

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

this takes care of both the protocol.

edited Apr 16, 2020 at 12:05

answered Apr 15, 2020 at 6:07

stanley oguazu

319 bronze badges

2 Comments

djibe Over a year ago

Indeed this module is provided with NodeJS.

Robin Métral Over a year ago

DO NOT USE. This is a legacy NodeJS API (check the docs). The newer WHATWG API is the same as in the browser: don't import "url" and just use a constructor: new URL(). It has been covered extensively in other answers. Finally, the question is about getting the hostname, but this answer just removes the path (it's not the same thing). Downvoted.

Babadinho · Accepted Answer · 2022-11-17 12:10:20Z

1

This solution works well and you can also use if URL contains a lot of invalid characters.

install psl package

npm install --save psl

implementation

const psl = require('psl');

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

console.log(parsed)

output:

{
  input: 'www.youtube.com',
  tld: 'com',
  sld: 'youtube',
  domain: 'youtube.com',
  subdomain: 'www',
  listed: true
}

answered Nov 17, 2022 at 12:10

Babadinho

431 silver badge10 bronze badges

1 Comment

Robin Métral Over a year ago

Same as stackoverflow.com/a/23945027, and psl is slow+heavy

Berthelot Loïc · Accepted Answer · 2023-04-18 17:39:34Z

1

Simple :

const url = new URL("https://www.magicspoon.com/pages/miss-cereal-new-bday");
domainUrl = url.hostname?.split(".").slice(-2).join(".");
//domainUrl: magicspoon.com
--- 
const url = new URL("https://magicspoon.com/pages/miss-cereal-new-bday");
domainUrl = url.hostname?.split(".").slice(-2).join(".");
//domainUrl: magicspoon.com

answered Apr 18, 2023 at 17:39

Berthelot Loïc

313 bronze badges

1 Comment

user67275 Over a year ago

Please add some explanation for your code rather than posting code only. Additional explanation will be more helpful.

Yeongjun Kim · Accepted Answer · 2016-09-09 02:59:59Z

0

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

edited Sep 9, 2016 at 2:59

answered Sep 8, 2016 at 8:48

Yeongjun Kim

7597 silver badges19 bronze badges

3 Comments

Kyle Strand Over a year ago

@ChristianTernus On the contrary; the OP mentioned regex, and this is pretty obviously a regex expression designed to match the requested portion of a URL. It's not entirely correct (e.g. it requires www. even though not all URLs have this component), but it is certainly an answer.

Christian Ternus Over a year ago

@KyleStrand Pretty obviously is a subjective judgement; providing a raw regex when asked "I'm seeking a JS/jQuery version of this solution" doesn't answer the qeustion.

Chamilyan Over a year ago

I'm the OP. I was a new developer at the time seeking an out of the box solution in JS. Indeed, a raw regex string without any context would not have helped at all. Plus it's incomplete.

Glen Thompson · Accepted Answer · 2020-05-12 23:13:04Z

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.

John Doherty · Accepted Answer · 2023-02-16 01:36:03Z

0

A URL is schema://domain/path/to/resource?key=value#fragment so you could split on /:

/**
 * Get root of URL
 * @param {string} url - string to parse
 * @returns {string} url root or empty string
 */
function getUrlRoot(url) {
  return String(url || '').split('/').slice(0, 3).join('/');
}

Example:

getUrlRoot('http://www.youtube.com/watch?v=ClkQA2Lb_iE');
// returns http://www.youtube.com

getUrlRoot('http://youtu.be/ClkQA2Lb_iE');
// returns http://youtu.be

getUrlRoot('http://www.example.com/12xy45');
// returns http://www.example.com

getUrlRoot('http://example.com/random');
// returns http://example.com

edited Feb 16, 2023 at 1:36

answered Feb 16, 2023 at 1:19

John Doherty

4,17540 silver badges40 bronze badges

Comments

MrMan · Accepted Answer · 2024-11-12 17:48:50Z

0

People why so complicated?

const ExtractDomain = url => url.includes('/') ? url.split('/')[2] : '';

answered Nov 12, 2024 at 17:48

MrMan

351 silver badge5 bronze badges

1 Comment

Community Over a year ago

Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center.

Chamilyan · Accepted Answer · 2019-09-22 22:41:55Z

-1

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

answered Sep 22, 2019 at 22:41

Chamilyan

9,45310 gold badges40 silver badges68 bronze badges

Comments

I_Tech · Accepted Answer · 2016-04-25 10:11:55Z

-6

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");
  }

answered Apr 25, 2016 at 10:11

I_Tech

11

1 Comment

piersadrian Over a year ago

OP was looking for an answer in JavaScript, not Java.

Collectives™ on Stack Overflow

30 Answers 30

1 Comment

1. Using npm package psl (Public Suffix List)

2. My Custom Implementation of extractRootDomain

3. URL(url).hostname

9 Comments

5 Comments

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1. Use URL.hostname for readability

2. Use RegEx for performance

TL;DR

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

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Linked

Related

2. My Custom Implementation of `extractRootDomain`

3. `URL(url).hostname`

1. Use `URL.hostname` for readability