Remove URL parameters JS ES6

Question

I have a url that contains a parameter, which I want to remove. I have created an if...else condition, but I would like to improve it in ES6.

if ( window.location.search.indexOf('&submit')  > -1 ) {
    window.location = urlBase + '/' + window.location.search;
} else {
    window.location = urlBase + '/' + window.location.search + '&submit=1';
}

Thank you in advance.

What would you hope that ES6 could improve in this code? – nils Mar 30 '17 at 14:09 — nils, Mar 30 '17 at 14:09
I want just know if possible to reduce this condition. – Kevin Py Mar 30 '17 at 14:10 — Kevin Py, Mar 30 '17 at 14:10

score 3 · Accepted Answer · answered Mar 30 '17 at 14:10

Not sure what your goal is, but here I've simply used includes and template literals.

if (window.location.search.includes('&submit')) {
    window.location = `${urlBase}/${window.location.search}`;
} else {
    window.location = `${urlBase}/${window.location.search}&submit=1`;
}

Remove URL parameters JS ES6

1 Answers1