What's the prettiest way to compare one value against multiple values?

Question

Whats the prettiest way to compare one value against multiples options?

I know there are loads of ways of doing this, but I'm looking for the neatest.

i ask because i'd hoped this was workable (it isn't, quite obviously when you look at it):

if (foobar == (foo||bar) ) {
     //do something
}

Answer 1

Don't try to be too sneaky, especially when it needlessly affects performance. If you really have a whole heap of comparisons to do, just format it nicely.

if (foobar === foo ||
    foobar === bar ||
    foobar === baz ||
    foobar === pew) {
     //do something
}

Answer 2

What i use to do, is put those multiple values in an array like

var options = [foo, bar];

and then, use indexOf()

if(options.indexOf(foobar) > -1){
   //do something
}

for prettiness:

if([foo, bar].indexOf(foobar) +1){
   //you can't get any more pretty than this :)
}

and for the older browsers:
( https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Array/IndexOf )

if (!Array.prototype.indexOf) {
    Array.prototype.indexOf = function (searchElement /*, fromIndex */ ) {
        "use strict";
        if (this == null) {
            throw new TypeError();
        }
        var t = Object(this);
        var len = t.length >>> 0;
        if (len === 0) {
            return -1;
        }
        var n = 0;
        if (arguments.length > 0) {
            n = Number(arguments[1]);
            if (n != n) { // shortcut for verifying if it's NaN
                n = 0;
            } else if (n != 0 && n != Infinity && n != -Infinity) {
                n = (n > 0 || -1) * Math.floor(Math.abs(n));
            }
        }
        if (n >= len) {
            return -1;
        }
        var k = n >= 0 ? n : Math.max(len - Math.abs(n), 0);
        for (; k < len; k++) {
            if (k in t && t[k] === searchElement) {
                return k;
            }
        }
        return -1;
    }
}

Answer 3

Since nobody has added the obvious solution yet which works fine for two comparisons, I'll offer it:

if (foobar === foo || foobar === bar) {
     //do something
}

And, if you have lots of values (perhaps hundreds or thousands), then I'd suggest making a Set as this makes very clean and simple comparison code and it's fast at runtime:

// pre-construct the Set
var tSet = new Set(["foo", "bar", "test1", "test2", "test3", ...]);

// test the Set at runtime
if (tSet.has(foobar)) {
    // do something
}

For pre-ES6, you can get a Set polyfill of which there are many. One is described in this other answer.

Answer 4

You can use a switch:

switch (foobar) {
  case foo:
  case bar:
    // do something
}

Answer 5

Just for kicks, since this Q&A does seem to be about syntax microanalysis, a tiny tiny modification of André Alçada Padez's suggestion(s):

(and of course accounting for the pre-IE9 shim/shiv/polyfill he's included)

if (~[foo, bar].indexOf(foobar)) {
    // pretty
}

Answer 6

Why not using indexOf from array like bellow?

if ([foo, bar].indexOf(foobar) !== -1) {
    // do something
}

Just plain Javascript, no frameworks or libraries but it will not work on IE < 9.

Answer 7

(foobar == foo || foobar == bar) otherwise if you are comparing expressions based only on a single integer, enumerated value, or String object you can use switch. See The switch Statement. You can also use the method suggested by André Alçada Padez. Ultimately what you select will need to depend on the details of what you are doing.

Answer 8

I like the pretty form of testing indexOf with an array, but be aware, this doesn't work in all browsers (because Array.prototype.indexOf is not present in old IExplorers).

However, there is a similar way by using jQuery with the $.inArray() function :

if ($.inArray(field, ['value1', 'value2', 'value3']) > -1) {
    alert('value ' + field + ' is into the list'); 
}

It could be better, so you should not test if indexOf exists.

Be careful with the comparison (don't use == true/false), because $.inArray returns the index of matching position where the value has been found, and if the index is 0, it would be false when it really exist into the array.