Reduce multiple ORs in IF statement in JavaScript

Question

Is there a simpler way to rewrite the following condition in JavaScript?

if ((x == 1) || (x == 3) || (x == 4) || (x == 17) || (x == 80)) {...}

Gumbo · Accepted Answer · 2010-04-13T15:37:16.260

17

You could use an array of valid values and test it with indexOf:

if ([1, 3, 4, 17, 80].indexOf(x) != -1)

Edit Note that indexOf was just added in ECMAScript 5 and thus is not implemented in every browser. But you can use the following code to add it if missing:

if (!Array.prototype.indexOf)
{
  Array.prototype.indexOf = function(elt /*, from*/)
  {
    var len = this.length >>> 0;

    var from = Number(arguments[1]) || 0;
    from = (from < 0)
         ? Math.ceil(from)
         : Math.floor(from);
    if (from < 0)
      from += len;

    for (; from < len; from++)
    {
      if (from in this &&
          this[from] === elt)
        return from;
    }
    return -1;
  };
}

Or, if you’re already using a JavaScript framework, you can also use its implementation of that method.

edited Apr 13 '10 at 15:37

answered Apr 13 '10 at 14:38

Gumbo

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1

+1... This is the neatest. The `in` operator does not work correctly for arrays, as you suggested: https://developer.mozilla.org/en/Core_JavaScript_1.5_Reference/Operators/Special_Operators/in_Operator – Daniel Vassallo Apr 13 '10 at 14:45
1

@sterofrog: You can add indexOf to older browsers: https://developer.mozilla.org/en/Core_JavaScript_1.5_Reference/Global_Objects/Array/indexOf – Daniel Vassallo Apr 13 '10 at 14:49
1

@Daniel: yes, this should be added to the answer, for gooogler's benefit – user187291 Apr 13 '10 at 15:07
1

Bear in mind that setting Array prototype members breaks the for (i in array) construct, as those prototype members will show up as items of every array. If Array prototype members are set (e.g. with this, by mootools, etc.), you have to iterate over the members with `for var i = 0; i < array.length; i++` , or use a framework function that handles it for you (e.g. jQuery.each()). – Joey Adams Apr 13 '10 at 15:42
1

@Joey Adams: `for … in` is for object properties and not array keys. *Always* use a the standard ranging `for` loop for array iteration. (jQuery does that too.) – Gumbo Apr 13 '10 at 15:54

score 7 · Answer 2 · 2010-04-14T14:20:38.383

7

switch (x) {
    case 1:
    case 3:
    case 4:
    case 17:
    case 80:
        //code
        break;
    default:
        //code
}

edited Apr 14 '10 at 14:20

answered Apr 13 '10 at 14:36

3

not exactly simpler, but a nice alternative, and good use of not using a break, which btw you forgot to put after the code in `case 80:` – thecoshman Apr 13 '10 at 14:37
1

and you need the break in the default case – BritishDeveloper Apr 13 '10 at 14:42
1

@BritishDeveloper the default clause doesn't require a break, as there cannot be any cases after the default. For example, see http://www.w3schools.com/js/js_switch.asp – cmptrgeekken Apr 13 '10 at 15:04
1

@cmptrgeekken true - i had my c# eyes in – BritishDeveloper Apr 13 '10 at 16:13

Edgar · Answer 3 · 2010-04-15T07:41:14.383

3

This is a little function I found somewhere on the web:

function oc(a) {
    var o = {};
    for (var i = 0; i < a.length; i++) {
        o[a[i]] = '';
    }
    return o;
}

Used like this:

if (x in oc(1, 3, 4, 17, 80)) {...}

I'm using it for strings myself; haven't tried with numbers, but I guess it would work.

edited Apr 15 '10 at 07:41

answered Apr 13 '10 at 15:24

Edgar

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score 2 · Answer 4 · answered Apr 13 '10 at 16:29

2

a regular expression test uses the string value of x:

if(/^[134]|17|80$/.test(x)){/*...*/}

answered Apr 13 '10 at 16:29

kennebec

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score 1 · Answer 5 · answered Apr 13 '10 at 14:44

1

You can optimize your own example and get rid of a few characters, making it easier on the eyes..:

if (x == 1 || x == 3 || x == 4 || x == 17 || x == 80) { ... }

answered Apr 13 '10 at 14:44

cllpse

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score 0 · Answer 6 · edited May 23 '17 at 12:06

0

many options

if ([0, 1, 3, 4, 17, 80].indexOf(x) > 0)

if(/^(1|3|4|17|80)$/.test(x))

if($.inArray(x, [1, 3, 4, 17, 80])

another one, based on Ed's answer

function list() {
    for (var i = 0, o = {}; i < arguments.length; i++)
        o[arguments[i]] = '';
    return o;
}


if(x in list(1, 3, 4, 17, 80))...

edited May 23 '17 at 12:06

Community

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answered Apr 13 '10 at 14:43

user187291

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JavaScript’s array index starts with 0. – Gumbo Apr 13 '10 at 14:44
@stereofrog: Yes: `[0, 1, 3, 4, 17, 80].indexOf(0) === 0`. – Gumbo Apr 13 '10 at 14:50
@stereofrog: so, I think `if ([0, 1, 3, 4, 17, 80].indexOf(x) > 0)` needs to be `if ([0, 1, 3, 4, 17, 80].indexOf(x) >= 0)` – Daniel Vassallo Apr 13 '10 at 15:34
@Daniel Vassallo, this would return 'true' for x=0, which is wrong – user187291 Apr 13 '10 at 15:51
@stereofrog: Ok, I see that you've added an extra element in the array. But why not `if ([1, 3, 4, 17, 80].indexOf(x) != -1)`? – Daniel Vassallo Apr 13 '10 at 16:00
@stereofrog: I promise I didn't downvote. But I still did not get why you prefer to add a fictitious first element. – Daniel Vassallo Apr 13 '10 at 17:50

score 0 · Answer 7 · answered Jun 17 '19 at 19:33

0

You can also use the Array.includes the simplest way...

    if([1,3,4,17,80].includes(x)){
        console.log(true); 
        // rest of the code
    }

answered Jun 17 '19 at 19:33

Santosh

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score 0 · Answer 8 · answered May 15 '20 at 08:44

0

Inspired by @Santosh I created a simplified version:

const input = x => [1, 3, 4, 17, 80].includes(x);

console.log(input(10));
console.log(input(1));

answered May 15 '20 at 08:44

Penny Liu

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Reduce multiple ORs in IF statement in JavaScript

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