Javascript compare 3 values

Question

I have 3 values that I want to compare f, g and h. I have some code to check that they all equal each other and that none of them are null. I've had a look online but could not find anything that seemed to answer my query. Currently I am checking the code in the following way...

if(g == h && g == f && f == h && g != null && f != null && h != null)
{
//do something
}

This is quite long winded and I might be adding more values, so I was just wondering if there is a quicker way to check that none of the values are null and that all the values equal each other?

Thanks in advance for any help.

You do not have to check all the variables if they are `!= null`.. If they are equal between them, you can check only one against `null`. The same with checking `f == h`.. if both `f` and `h` are equal to `g` you do not have to check those two .. — Gabriele Petrioli, Apr 02 '12 at 08:50
have you compared the trade offs of putting all the var into an array? — cctan, Apr 02 '12 at 08:50
If first three equations will be true then you must check only one variable to test if it is not null so: — Bartek Kobyłecki, Apr 02 '12 at 08:51
Yes I could put them into an array, would it be easier to check them all if they were in an array? — Phil, Apr 02 '12 at 08:54

Gabriele Petrioli · Accepted Answer · 2023-03-20T20:38:04.953

64

You could shorten that to

if(g === h && g === f && g !== null)
{
//do something
}

For an actual way to compare multiple values (regardless of their number)
(inspired by/ simplified @Rohan Prabhu answer)

function areEqual(){
   var len = arguments.length;
   for (var i = 1; i< len; i++){
      if (arguments[i] === null || arguments[i] !== arguments[i-1])
         return false;
   }
   return true;
}

and call this with

if( areEqual(a,b,c,d,e,f,g,h) )
{
//do something
}

And a more recent approach

function areEqualNotNull(values) {
  if (values[0] === null) return false;

  return values.every(value => value === values[0]);
}

edited Mar 20 '23 at 20:38

answered Apr 02 '12 at 08:52

Gabriele Petrioli

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You could even omit the `!= null` – m90 Apr 02 '12 at 08:54
@m90, why ? what if all are null ? – Gabriele Petrioli Apr 02 '12 at 08:55
@m90 buy what if all of them are null? then it will evaluate to `true`. – David Hellsing Apr 02 '12 at 08:55
If all of them are null `Boolean(g)` will be `false`: http://jsfiddle.net/SFDRK/ Or am I missing something? – m90 Apr 02 '12 at 08:58
@m90.. ah ok.. i thought you meant to remove the whole `g != null` part.. But if you are comparing boolean values and they are all false ? it should return true according to specs.. – Gabriele Petrioli Apr 02 '12 at 09:02
2

@m90 true, but that also goes for all falsy values, the OP’s asked specifically for `null` http://jsfiddle.net/SFDRK/1/ – David Hellsing Apr 02 '12 at 09:02

score 35 · Answer 2 · answered Nov 19 '20 at 20:18

35

What about

(new Set([a,b,c])).size === 1

answered Nov 19 '20 at 20:18

olidem

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4

That's definitely a very elegant solution that I'd never have considered. I wonder if it has any disadvantages. – Astridax Dec 12 '20 at 14:15
2

Nice solution! keep in mind, @Phil, that 1(Number) and "1"(String) will be treated as different values – Vlad R Mar 12 '21 at 06:32
Very elegant. The size property of Set returns the number of unique elements in a Set object. – Gene Myers Feb 13 '22 at 11:12
@Astridax disadvantages are that it allocates and garbage collects two objects (slow) and doesn't short circuit (also slow). But might be premature optimization not to use it if it's cleanest. – ggorlen May 14 '22 at 15:11
Also, OP says "none of them are null" but `new Set([null, null, null]).size === 1` is true, so this doesn't really fulfill the specification. That said, not being sure whether numbers are null probably indicates a deeper design problem. – ggorlen May 14 '22 at 15:16

score 20 · Answer 3 · answered Sep 25 '15 at 19:01

20

Works for any number of items.

ES5

if ([f, g, h].every(function (v, i, a) {
  return (
    v === a[0] &&
    v !== null
  );
})) {
  // Do something
}

ES2015

if ([f, g, h].every((v, i, a) => 
  v === a[0] &&
  v !== null
)) {
  // Do something
}

answered Sep 25 '15 at 19:01

JMM

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3
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score 5 · Answer 4 · answered Apr 02 '12 at 08:52

I suggest you write a function where you give an array with all the values you want to compare and then iterate through the array to compare the values which each other:

function compareAllValues(a) {
     for (var i = 0; i < a.length; i++) {
         if (a[i] === null) { return false }
         for (var j = 0; j < i; j++) {
            if (a[j] !== a[i]) { return false }
         }
     }

     return true;
}

that should be it, I think :)

Your soln. is much space complex than it should be. – Rajesh Paul Oct 18 '13 at 04:57 — Rajesh Paul, Oct 18 '13 at 04:57

score 4 · Answer 5 · answered Apr 02 '12 at 08:52

Write a simple function:

var checkAllArguments = function() {
    var len = arguments.length;
    var obj;

    if(len == 0) {
        return true;
    } else {
        if(arguments[0] == null) {
            return false;
        } else {
            obj = arguments[0];
        }
    }

    for(var i=1; i<len; i++) {
        if(arguments[i] == null) {
            return false;
        }

        if(obj == arguments[i]) {
            continue;
        } else {
            return false;
        }
    }

    return true;
}

Now, to check multiple arguments, all you have to do is:

if(checkAllArguments(g, h, f)) {
   // Do something
}

score 0 · Answer 6 · answered Sep 30 '19 at 17:16

Create an array of string and check the existance of next value in there

compareAnswers(a1: string, a2: string, a3: string): boolean {
   this.compareAnswerArr.push(a1);
if (this.compareAnswerArr.includes(a2) == false) {
  this.compareAnswerArr.push(a2);
  if (this.compareAnswerArr.includes(a3) == false) {
    return true;
  } else return false;
}
else return false;}

score -2 · Answer 7 · answered Dec 07 '16 at 13:27

-2

Add a distinct function to Array.prototype:

Array.prototype.distinct = function() {
    var result = [];
    for(var i = 0; i < this.length; i++) {
        if (result.indexOf(this[i]) == -1) {
            result.push(this[i]);
        }
    }
    return result; 
}

Then do:

var myArray = [f, g, h];
if (myArray.indexOf(null) == -1 && myArray.unique().length == 1)
{
    // no nulls and all elements have the same value!
}

answered Dec 07 '16 at 13:27

Kris

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Note that I no longer recommend this – Kris Sep 19 '21 at 21:27
Thanks for the note, but if you no longer recommend it, consider [edit]ing it to something you do recommend or deleting it altogether. It's perfectly fine for answers to change as best practices move on. – ggorlen May 14 '22 at 15:14
You also have a typo: `unique()` rather than `distinct()`. Best to use `===` rather than `==`. – ggorlen May 14 '22 at 15:19

Javascript compare 3 values

7 Answers7

ES5

ES2015

Linked

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