How to check if a variable equals one of two values if a clean manner in JS

Question

To check if x variable is equal to 1 or 2, i would do normally something like:

if (x === 1 || x === 2){
   // ....
}

but this can get in some cases very cumbersome.

Edit :I'm working right now on this and writing the fonction name every time i think can be done in a cleaner manner:

if (
      this.getNotificationStatus() === 'denied' ||
      this.getNotificationStatus() === 'blocked'
    )

Is there any other lighter way to write this?

THANKS

If the comparison involves only two values, then the `if` loop provides better readability — Krishna Prashatt, Jan 31 '20 at 09:30
Note that [Array.prototype.includes() tests reference equality](https://codesearchable.com/it/4531896/) — Grey Chanel, Jan 31 '20 at 09:43

score 2 · Accepted Answer · answered Jan 31 '20 at 09:30

2

You could do:

if ([1, 2].includes(x)) {
  // ....
}

Or:

if ([1, 2].indexOf(x) > -1) {
  // ....
}

Or:

switch (x) {
  case 1:
  case 2:
    // ....
    break;
  default:
}

I don't think they're "lighter" than your solution though.

answered Jan 31 '20 at 09:30

technophyle

Note: `includes` doesn't work prior ES2016. `indexOf` doesn't supported in some old browsers (IE9 for example). – NoSkill Jan 31 '20 at 09:37
Yes, and `includes` doesn't work in IE at all. – technophyle Jan 31 '20 at 09:37

score 0 · Answer 2 · answered Jan 31 '20 at 09:29

0

Try this:

if ([1, 2, 3].includes(x)) {
  // your code here
}

answered Jan 31 '20 at 09:29

Grey Chanel

2 Answers2