4

I have a URL with a query string which starts and ends with a specific letters. Here's an example and my approach:

Let's say the url in the address bar is "http://localhost:3001/build/?videoUrl=bitcoin.vid.com/money#/"

I first extract this url using the window.location.href and save it in a variable x.

Now, I want to first check if the videoUrl is present in the URL or not and then if it is available, I split the URL and extract the required URL which is bitcoin.vid.com/money

let x = "http://localhost:3001/build/?videoUrl=bitcoin.vid.com/money#/";
let y;
let result;
if(x.indexOf("?videoUrl")>-1) {
   y = x.split("?videoUrl=");
   result = y[1].split("#")[0];
   console.log("Resultant URL:", result);
}

I feel the whole code which I wrote is a bit cumbersome. Could anyone let me know if there is more elegant way to do the same?

Note: videoUrl is not available always in the URL, hence the check if it exists. And please also let me know if I need to further make any checks?

Alicia Sykes
  • 5,997
  • 7
  • 36
  • 64
Angela Roux
  • 473
  • 1
  • 5
  • 14
  • Possible duplicate of [How can I get query string values in JavaScript?](https://stackoverflow.com/questions/901115/how-can-i-get-query-string-values-in-javascript) – Herohtar Jan 04 '18 at 18:57
  • 2
    - https://developer.mozilla.org/en-US/docs/Web/API/URLSearchParams - https://caniuse.com/#search=URLSearchParams – Peter Seliger Jan 04 '18 at 19:03

2 Answers2

11

Your method works, which is the main thing- but for cleaner and more robust approach, you can make use of the built-in JavaScript URL() class.

let x = "http://localhost:3001/build/?videoUrl=bitcoin.vid.com/money#/";
console.log((new URL(x)).searchParams.get("videoUrl"));

If your videoUrl parameter is not present, then this will just return null, and you can deal with that accordingly


If you want to make this slightly more reusable, here's a function that does what you need:

const getParamFromUrl = (url, param) =>
    (new URL(url)).searchParams.get(param);

There's plenty more information about implementation and compatibility of this in the documentation, here: https://developer.mozilla.org/en-US/docs/Web/API/URL

Alicia Sykes
  • 5,997
  • 7
  • 36
  • 64
2

Good question!

Two options for a neat solution:

  • Try using the URL approach thatr @Lissy suggested, the functionality to do what you need is all built in. However she didn't mention the potential flaws when it comes to corss-browser functionality
  • Alternativley, I'd just use a simple regex, e.g. (\?|\&)([^=]+)\=([^&]+) - in my opinion, it is much simpler