How can I delete a query string parameter in JavaScript?

You can change the URL with:

window.history.pushState({}, document.title, window.location.pathname);

this way, you can overwrite the URL without the search parameter, I use it to clean the URL after take the GET parameters.

I don't see major issues with a regex solution. But, don't forget to preserve the fragment identifier (text after the #).

Here's my solution:

function RemoveParameterFromUrl(url, parameter) {
  return url
    .replace(new RegExp('[?&]' + parameter + '=[^&#]*(#.*)?$'), '$1')
    .replace(new RegExp('([?&])' + parameter + '=[^&]*&'), '$1');
}

And to bobince's point, yes - you'd need to escape . characters in parameter names.

Modern browsers provide URLSearchParams interface to work with search params. Which has delete method that removes param by name.

if (typeof URLSearchParams !== 'undefined') {
  const params = new URLSearchParams('param1=1&param2=2&param3=3')
  
  console.log(params.toString())
  
  params.delete('param2')
  
  console.log(params.toString())

} else {
  console.log(`Your browser ${navigator.appVersion} does not support URLSearchParams`)
}

"[&;]?" + parameter + "=[^&;]+"

Seems dangerous because it parameter ‘bar’ would match:

?a=b&foobar=c

Also, it would fail if parameter contained any characters that are special in RegExp, such as ‘.’. And it's not a global regex, so it would only remove one instance of the parameter.

I wouldn't use a simple RegExp for this, I'd parse the parameters in and lose the ones you don't want.

function removeURLParameter(url, parameter) {
    //prefer to use l.search if you have a location/link object
    var urlparts = url.split('?');   
    if (urlparts.length >= 2) {

        var prefix = encodeURIComponent(parameter) + '=';
        var pars = urlparts[1].split(/[&;]/g);

        //reverse iteration as may be destructive
        for (var i = pars.length; i-- > 0;) {    
            //idiom for string.startsWith
            if (pars[i].lastIndexOf(prefix, 0) !== -1) {  
                pars.splice(i, 1);
            }
        }

        return urlparts[0] + (pars.length > 0 ? '?' + pars.join('&') : '');
    }
    return url;
}