remove hostname and port from url using regular expression

Question

I am trying to remove

http://localhost:7001/

part from

http://localhost:7001/www.facebook.com

to get the output as

www.facebook.com

what is the regular expression that i can use to achieve this exact pattern?

Answer 1

You don't need any library or REGEX

var url = new URL('http://localhost:7001/www.facebook.com')
console.log(url.pathname)

https://developer.mozilla.org/en-US/docs/Web/API/URL

Answer 2

Based on @atiruz answer, but this is

url = url.replace( /^[a-zA-Z]{3,5}\:\/{2}[a-zA-Z0-9_.:-]+\//, '' );

• shortest
• can take https or ftp too
• can take url with or without explicit port

Answer 3

To javascript you can use this code:

var URL = "http://localhost:7001/www.facebook.com";
var newURL = URL.replace (/^[a-z]{4,5}\:\/{2}[a-z]{1,}\:[0-9]{1,4}.(.*)/, '$1'); // http or https
alert (newURL);

Look at this code in action Here

Regards, Victor

Answer 4

This is how I made it work without resorting to regular expressions:

var URL = "http://localhost:7001/www.facebook.com";

var URLsplit = URL.split('/');

var host = URLsplit[0] + "//" + URLsplit[2] + "/";

var newURL = URL.replace(host, '');

Might not be an elegant solution though but it should be easier to understand for those who don't have much experience with regex (like me! ugh!).

Answer 5

For a simple regex to match any protocol, domain, and (optionally) port:

var url = 'http://localhost:7001/www.facebook.com';

// Create a regex to match protocol, domain, and host
var matchProtocolDomainHost = /^.*\/\/[^\/]+:?[0-9]?\//i;

// Replace protocol, domain and host from url, assign to `myNewUrl`
var myNewUrl = url.replace(matchProtocolDomainHost, '');

Now myNewUrl === 'www.facebook.com'.

See demo on regex101

Answer 6

Regex to match the part of url, that you want to remove, will be something like: /^http[s]?:\/\/.+?\//

Example of Java code (note that in Java we use two backslashes "\\" for escaping character):

String urlWithBasePath = "http://localhost:7001/www.facebook.com";
String resultUrl = urlWithBasePath.replaceFirst("^http[s]?:\\/\\/.+?\\/", ""); // resultUrl => www.facebook.com

Example of JS code:

let urlWithBasePath = "http://localhost:7001/www.facebook.com";
let resultUrl = urlWithBasePath.replace(/^http[s]?:\/\/.+?\//, ''); // resultUrl => www.facebook.com

Example of Python code:

import re
urlWithBasePath = "http://localhost:7001/www.facebook.com"
resultUrl = re.sub(r'^http[s]?:\/\/.+?\/', '', urlWithBasePath) # resultUrl => www.facebook.com

Example or Ruby code:

urlWithBasePath = "http://localhost:7001/www.facebook.com"
resultUrl =  urlWithBasePath = urlWithBasePath.sub(/^http[s]?:\/\/.+?\//, '') # resultUrl => www.facebook.com

Example of PHP code:

$urlWithBasePath = "http://localhost:7001/www.facebook.com";
$resultUrl = preg_replace('/^http[s]?:\/\/.+?\//', '', $urlWithBasePath); // resultUrl => www.facebook.com

Example of C# code (you should also specify using System.Text.RegularExpressions;):

string urlWithBasePath = "http://localhost:7001/www.facebook.com";
string resultUrl = Regex.Replace(urlWithBasePath, @"^http[s]?:\/\/.+?\/", ""); // resultUrl => www.facebook.com

Answer 7

All other regular expressions here look a bit complicated? This is all that's needed: (right?)

var originSlash = /^https?:\/\/[^/]+\//i;

theUrl.replace(originSlash, '');

Answer 8

Alternatively, you can parse the url using as3corelib's URI class. That way you don't have to do any string manipulations, which helps to avoid making unintentional assumptions. It requires a few more lines of code, but it's a more general solution that should work for a wide variety of cases:

var url : URI = new URI("http://localhost:7001/myPath?myQuery=value#myFragment");

// example of useful properties
trace(url.scheme); // prints: http
trace(url.authority); // prints the host: localhost
trace(url.port); // prints: 7001
trace(url.path); // prints: /myPath
trace(url.query); // prints: myQuery=test
trace(url.fragment); // prints: myFragment

// build a new relative url, make sure we keep the query and fragment
var relativeURL : URI = new URI();
relativeURL.path = url.path;
relativeURL.query = url.query;
relativeURL.fragment = url.fragment;

var relativeURLString : String = relativeURL.toString();

// remove first / if any
if (relativeURLString.charAt(0) == "/") {
    relativeURLString = relativeURLString.substring(1, relativeURLString.length);
}

trace(relativeURLString); // prints: myPath?myQuery=test#myFragment

Answer 9

instead of using regex you could just use the browser's capabilities of parsing an URL:

var parser = document.createElement('a');
parser.href = "http://localhost:7001/www.facebook.com";
var path = parser.pathname.substring(1); // --> results in 'www.facebook.com'

Answer 10

If you are just looking to remove the origin and get the rest of the URL, including hashes, query params and any characters without restrictions:

function getUrlFromPath(targetUrl) {
  const url = new URL(targetUrl);
  return targetUrl.replace(url.origin, '');
}

function main() {
  const testUrls = [
    'http://localhost:3000/test?search=something',
    'https://www.google.co.in/search?q=hello+there+obi+wan&newwindow=1&sxsrf=ALiCzsZoaZvs0CrLQEHFmmR-MdrZ2ZHW2A%3A1665462761920&source=hp&ei=6fFEY_7cNY36wAOFyqagBA&iflsig=AJiK0e8AAAAAY0T_-R12vR7P_tmmkpEqgzmoZNczbnZA&ved=0ahUKEwi-9buirNf6AhUNPXAKHQWlCUQQ4dUDCAc&uact=5&oq=hello+there+obi+wan&gs_lcp=Cgdnd3Mtd2l6EAMyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEOgQIIxAnOhEILhCABBCxAxCDARDHARDRAzoLCAAQgAQQsQMQgwE6CwguEIAEELEDEIMBOg4ILhCABBCxAxCDARDUAjoICAAQsQMQgwE6CwguEIAEELEDENQCOggIABCABBCxAzoICC4QsQMQgwFQAFjjE2C6FmgAcAB4A4AB1QSIAd8ZkgELMC45LjIuMC4yLjGYAQCgAQE&sclient=gws-wiz'
  ];
  testUrls.forEach(url => {
    console.log(getUrlFromPath(url));
  });
}

main();

A failsafe regex pattern to achieve this will get complex and cumbersome to come up with.

Answer 11

Just use replace

"http://localhost:7001/www.facebook.com".replace("http://localhost:7001/",'')