JS - array.includes issue

Question

I have this issue with a JS function array.includes. I have this array:

The thing is when I use this code, nothing will happen.

var array_type;
//array has these 2 values:
//array_type[0] == 0;
//array_type[1] == 2;
if (array_type.includes(2)) {
 console.log("good");
}

Do you have any idea why? Thank you for any help.

[0, 2].includes(2) === true here. You'll need to provide more code to demonstrate your issue. — qubyte
– qubyte, Commented Feb 22, 2018 at 16:20
You have to show us where you initialise array_type and where you use it — theAlexandrian
– theAlexandrian, Commented Feb 22, 2018 at 16:23

Sanchit Patiyal · Accepted Answer · 2018-02-22 16:35:38Z

4

If you are using Internet Explorer then array.includes() will not work. Instead you need to use indexOf. Internet Explorer doesn't have a support for Array.includes()

var array_type = [0, 2];

if (array_type.indexOf(2) !== -1) {
  console.log("good");
}

References for includes()

References for indexOf()

Check the browser compatibility sections in the link

edited Feb 22, 2018 at 16:35

answered Feb 22, 2018 at 16:25

Sanchit Patiyal

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1 Comment

Sanchit Patiyal Over a year ago

@AdamŠulc Glad I helped :)

juan garcia · Accepted Answer · 2018-02-22 16:21:04Z

2

This code works

[1,2].includes(2)

but you have to be careful if you can use the includes function

https://caniuse.com/#search=includes

answered Feb 22, 2018 at 16:21

juan garcia

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2 Comments

Adam Šulc Over a year ago

I did not know that I have to put [0,1] in there. Is there any way that it woud work without it? Because not everytime the array will contain only two items.

juan garcia Over a year ago

yes of course you can use functions of array like push / unshift to provide values to the array. developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

cbrawl · Accepted Answer · 2018-02-22 16:23:24Z

2

The code works for me. For example,

var array_type = [0, 2];

if (array_type.includes(2)) {
    console.log("good");
}

will log good.

Make sure you are properly inserting the items into the array.

answered Feb 22, 2018 at 16:23

cbrawl

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Comments

aconsuegra · Accepted Answer · 2018-02-22 16:27:34Z

2

Here your are not adding values, you are testing if array_type[0] is equal to 0 and array_type[1] is equal to 2

//array has these 2 values:
array_type[0] == 0;
array_type[1] == 2;

So this code

if (array_type.includes(2)) {
    console.log("good");
}

never be true Try

var array_type = [];
array_type[0] = 0;
array_type[1] = 2;

if (array_type.includes(2)) {
    console.log("good");
}

answered Feb 22, 2018 at 16:27

aconsuegra

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Comments

L Y E S - C H I O U K H · Accepted Answer · 2018-02-22 16:23:58Z

1

var array_type = [];
//array has these 2 values:
array_type[0] = 0;
array_type[1] = 2;
if (array_type.includes(2)) {
     console.log("good");
}

This should work!

answered Feb 22, 2018 at 16:23

L Y E S - C H I O U K H

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Comments

Ignacio Ara · Accepted Answer · 2018-02-22 16:27:17Z

1

I suppose that both comments is a value set (=) instead of a comparison (==)

Because using first option, it works:

> array_type.includes(2)
true

answered Feb 22, 2018 at 16:27

Ignacio Ara

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