How to check with javascript if connection is local host?

Question

I want to have a check in my javascript if the page loading up is on my local machine.

The reason why I want to do that is that when I developing I like to make sure that both my server side(C#) validation is working correctly. So I like to see both the client side and server sides errors to show up.

So while I am testing I have a flag in my jquery validate stuff that just always lets invalid data go through. This way I see the client side and server errors at one go.

However right now I have to manually go and change back and forth when going from development to production.

Answer 1

The location.hostname variable gives you the current host. That should be enough for you to determine which environment you are in.

if (location.hostname === "localhost" || location.hostname === "127.0.0.1")
    alert("It's a local server!");

Answer 2

if launching static html in browser, eg from location like file:///C:/Documents and Settings/Administrator/Desktop/ detecting "localhost" will not work. location.hostname will return empty string. so

if (location.hostname === "localhost" || location.hostname === "127.0.0.1" || location.hostname === "")
    alert("It's a local server!");

Answer 3

That's how it get checked in React, register service worker, good way to check if you are on localhost by checking hostname, including localhost and IPv6, and matching start with 127:

const isLocalhost = Boolean(
    window.location.hostname === 'localhost' ||
    // [::1] is the IPv6 localhost address.
    window.location.hostname === '[::1]' ||
    // 127.0.0.1/8 is considered localhost for IPv4.
    window.location.hostname.match(
        /^127(?:\.(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)){3}$/
    )
);

Answer 4

Still not a catch all but it might be a little improvement. You can now create an array of domains and use .includes

const LOCAL_DOMAINS = ["localhost", "127.0.0.1", ...];

if (LOCAL_DOMAINS.includes(window.location.hostname))
  alert("It's a local server!");

Answer 5

The following code checks whether an address is localhost, that is, whether it points to the same device from which this request is made.

export function isLocalHost(hostname = window.location.hostname) {
  return ['localhost', '127.0.0.1', '', '::1'].includes(hostname)
}

The following code also checks if the request goes into the local network. Some common cases where local network IPs start with 10.0. or 192.168. or mDNS / Bonjour like domains ending on .local:

export function isLocalNetwork(hostname = window.location.hostname) {
  return (
    (['localhost', '127.0.0.1', '', '::1'].includes(hostname))
    || (hostname.startsWith('192.168.'))
    || (hostname.startsWith('10.'))
    || (hostname.endsWith('.local'))
  )
}

Answer 6

Shortest form using same mechanic as other scripts:

if ( ["localhost", "127.0.0.1", ""].includes(window.location.hostname) ) {
     console.log("It's local host !");
}

Answer 7

const LOCAL_DOMAINS = [ "localhost", "127.0.0.1" ];

/* offline || development */
if ( LOCAL_DOMAINS.includes(location.hostname) )
{
    BASE_URL_PUBLIC = location.hostname + "/folder/website/"; // your project folder
}

/* online || production */
else
{
    BASE_URL_PUBLIC = location.hostname;
}

Answer 8

An easy way to do this would be to just check the hostname against localhost or check your custom domain name against a substring, in this case ".local" urls, such as http://testsite.local

var myUrlPattern = '.local';
if (window.location.hostname === "localhost" || location.hostname === "127.0.0.1" || window.location.hostname.indexOf(myUrlPattern) >= 0) {
    alert("It's a local server!");
}

Answer 9

You could detect in one of your code behind pages with c#, like this:

if ((Request.Url.Host.ToLower() == "localhost"))
{
    // ..., maybe set an asp:Literal value that's in the js
}

Or if you want to do it from client script, you could check the value of window.location.host.

if (window.location.host == "localhost")
{
    // Do whatever
}

Hope this helps.

Answer 10

The above answers mostly solve the problem but...

• What if localhost isn't necessarily 'localhost/'?
• What if you want to do FE validation during development?
• What if you want different behaviors during dev
  (fe validation, be validation, no validation)

One solution is to set the location hash and check it.

http://myname.foo.com/form.html#devValidation

You could add unlimited options with a switch

switch(location.hash) {}
    case '#devValidation':
        // log the results and post the form
        break;
    case '#beValidation':
        // skip front end validation entirely
        break;
    case '#noValidation':
        // skip all validation $('[name=validationType']).val('novalidation');
        break;
    case '#feValidation':
    default:
        // do fe validation
        break;
}

Answer 11

Regular expression is slower*, but short and neat. Also, nobody here checks for IPv6 localhost (::1)

/localhost|127\.0\.0\.1|::1|\.local|^$/i.test(location.hostname)

It checks for general localhost, .local domain and file: (empty hostname).

*) In Chrome, performance of [].includes(...) is the best (42 ms), followed by simple loop (for, while) with array item checking (119 ms), then [].indexOf(...) > -1 (289 ms) and finally the regexp (566 ms). But those measurements are somehow relative, because different browsers are optimized differently. In FF 52 ESR includes and indexOf have similar results, regexp is 2× slower and loop 6× slower.

Answer 12

Based on the above comments th following regular expression has helped me to verify if the url is 'localhost', any IP adress IPv4 or IPv6.

window.location.hostname.match(/localhost|[0-9]{2,3}\.[0-9]{2,3}\.[0-9]{2,3}\.[0-9]{2,3}|::1|\.local|^$/gi)

Answer 13

Many of the answers here miss some edge-cases. For example, depending on DNS/hosts/etc, you could have something.localhost.

So here is a slightly more comprehensive ES6 answer:

const localHost = ['localhost', '127.0.0.1', '::1', ''].includes(window.location.hostname) || window.location.hostname.endsWith('.localhost')