Get protocol, domain, and port from URL

Get protocol, domain, and port from URL

I need to extract the full protocol, domain, and port from a given URL. For example:
https://localhost:8181/ContactUs-1.0/contact?lang=it&report_type=consumer
>>>
https://localhost:8181

Solutions/Answers:

Solution 1:

first get the current address

var url = window.location.href

Then just parse that string

var arr = url.split("/");

your url is:

var result = arr[0] + "//" + arr[2]

Hope this helps

Solution 2:

var full = location.protocol+'//'+location.hostname+(location.port ? ':'+location.port: '');

Solution 3:

None of these answers seem to completely address the question, which calls for an arbitrary url, not specifically the url of the current page.

Method 1: Use the URL API (caveat: no IE11 support)

You can use the URL API (not supported by IE11, but available everywhere else).

This also makes it easy to access search params. Another bonus: it can be used in a Web Worker since it doesn’t depend on the DOM.

const url = new URL('http://example.com:12345/blog/foo/bar?startIndex=1&pageSize=10');

Method 2 (old way): Use the browser’s built-in parser in the DOM

Use this if you need this to work on older browsers as well.

//  Create an anchor element (note: no need to append this element to the document)
const url = document.createElement('a');
//  Set href to any path
url.setAttribute('href', 'http://example.com:12345/blog/foo/bar?startIndex=1&pageSize=10');

That’s it!

The browser’s built-in parser has already done its job. Now you can just grab the parts you need (note that this works for both methods above):

//  Get any piece of the url you're interested in
url.hostname;  //  'example.com'
url.port;      //  12345
url.search;    //  '?startIndex=1&pageSize=10'
url.pathname;  //  '/blog/foo/bar'
url.protocol;  //  'http:'

Bonus: Search params

Chances are you’ll probably want to break apart the search url params as well, since ‘?startIndex=1&pageSize=10’ isn’t too useable on its own.

If you used Method 1 (URL API) above, you simply use the searchParams getters:

url.searchParams.get('startIndex');  // '1'

Or to get all parameters:

Array
    .from(url.searchParams)
    .reduce((accum, [key, val]) => {
        accum[key] = val;
        return accum;
    }, {});
// -> { startIndex: '1', pageSize: '10' }

If you used Method 2 (the old way), you can use something like this:

// Simple object output (note: does NOT preserve duplicate keys).
var params = url.search.substr(1); // remove '?' prefix
params
    .split('&')
    .reduce((accum, keyval) => {
        const [key, val] = keyval.split('=');
        accum[key] = val;
        return accum;
    }, {});
// -> { startIndex: '1', pageSize: '10' }

Solution 4:

For some reason all the answers are all overkills. This is all it takes:

window.location.origin

More details can be found here: https://developer.mozilla.org/en-US/docs/Web/API/window.location#Properties

Solution 5:

As has already been mentioned there is the as yet not fully supported window.location.origin but instead of either using it or creating a new variable to use, I prefer to check for it and if it isn’t set to set it.

For example;

if (!window.location.origin) {
  window.location.origin = window.location.protocol + "//" + window.location.hostname + (window.location.port ? ':' + window.location.port: '');
}

I actually wrote about this a few months back A fix for window.location.origin

Solution 6:

host

var url = window.location.host;

returns localhost:2679

hostname

var url = window.location.hostname;

returns localhost