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

Answer 1

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!
}

Answer 2

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;}

Answer 3

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
}

Answer 4

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 :)

Answer 5

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
}

Answer 6

What about

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