Javascript Url Remove All Query Params Except

Jul 27, 2021
const urlObj = new URL('')// remove all query params except x and yconst keys = Array.from(urlObj.searchParams.keys())keys.forEach(key => {  if (!(['x', 'y'].includes(key))) {    urlObj.searchParams.delete(key)  }})  let url = urlObj.toString() //"

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✨ By Desmond Lua

A dream boy who enjoys making apps, travelling and making youtube videos. Follow me on @d_luaz

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