How to compare length of three array? Javascript

Question

I need to return false or alert or console.log('no true') if three list there is not the same number of items?

example
arrOne = [1,2,3];
arrTwo = [5,6,7];
arrTthree = [1,9];

FALSE arrThree has 2 values , one and two have 3 values. If same return true.

`arrOne.length === arrTwo.length && arrTwo.length === arrTthree.length` — 001, Nov 10 '20 at 01:40
Does this answer your question? [How to compare arrays in JavaScript?](https://stackoverflow.com/questions/7837456/how-to-compare-arrays-in-javascript) — geofh, Nov 10 '20 at 01:42
@geofh They just want to compare the lengths, not the contents. — Barmar, Nov 10 '20 at 01:44

Jay · Answer 1 · 2020-11-10T02:21:26.947

1

I have similar idea like @geofh, but it need not wrap all your arrays into another one as input.

function compare(...arrays) {
  return arrays.every(arr => arr.length === arrays[0].length);
}

With it, you could invoke compare([1, 2, 3], [5, 6, 7], [1, 9], ...)

edited Nov 10 '20 at 02:21

answered Nov 10 '20 at 02:15

Jay

34
4

score 0 · Answer 2 · answered Nov 10 '20 at 01:50

OK, I owe you an answer since I erroneously marked it as duplicate. To generalize, you need a sameLength function that takes an array of arrays, perhaps. That could look like:

function sameLength(arrays) {
    return arrays.slice(1).every(arr => arr.length === arrays[0].length)
}

so sameLength([arr1, arr2, arr3]) would return false in your case.

How to compare length of three array? Javascript

2 Answers2