Javascript array null check without iterating

Question

Hi I have an array like: [null,null,null,null] in javascript. How can I know if the array has all values equal to null without iterating over it?

just checking if there is a JS trick or shortcut to identify null array. — Aditya Kappagantula
– Aditya Kappagantula, Commented Sep 9, 2015 at 0:50
Avoiding iteration is impossible. If you don't loop over the array yourself, whatever method you call will have to. Is that your goal, to avoid explicitly looping over it yourself? That's possible. — John Kugelman
– John Kugelman, Commented Sep 9, 2015 at 0:50
Yep that is my goal. To avoid manual iteration. Code look ugly if i do that. — Aditya Kappagantula
– Aditya Kappagantula, Commented Sep 9, 2015 at 0:51
@Brendan - Or better yet for(var i = 0; i < arr.length; i++) if(array[i] !== null) allNull = false; — PitaJ
– PitaJ, Commented Sep 9, 2015 at 1:04

PitaJ · Accepted Answer · 2015-09-09 01:08:08Z

6

There is no way to check for all null values, you would have to iterate over it, or the function you call will have to in the background. However, if you want, you could use the native Array method filter to help you out.

var allNull = !arr.filter(Boolean).length;

This will work with any falsy value, like undefined, zero, NaN, or an empty string. A more precise form is

var allNull = !arr.filter(function(elem){ return elem !== null; }).length;

Which would check for only null.

Another possibility is using .join:

var allNull = !arr.join("").length;

This checks for null, undefined, or ""

edited Sep 9, 2015 at 1:08

answered Sep 9, 2015 at 0:52

PitaJ

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Comments

RobG · Accepted Answer · 2015-09-09 00:55:02Z

4

There are various built–in methods to help, one is every to test that every value isn't null. It will return false at the first null:

var noNulls = array.every(function(value){return value !== null});

Another is some, which will return true at the first null:

var hasNulls = array.some(function(value){return value === null});

answered Sep 9, 2015 at 0:55

RobG

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1 Comment

PitaJ Over a year ago

every is indeed a helpful method. I didn't even remember it existed.

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