jQuery check in array and return true or false

Question

I have an array of pages

var $pages = ["events", "template", "privacy", "terms", "mentor", "party", "getinvolved", "tools"];

What I want to do is

  if ( $body[0].id !== anyOfThePageInTheArray ) {
    //do something here
    });

How can I do like if the page that they are on is not one of the page that is in the array.

I have tried .inArray and .each but I think maybe I'm doing it wrong or something.

Blender · Accepted Answer · 2013-06-13 19:43:32Z

2

The native Array.indexOf works fine:

if ($pages.indexOf($body[0].id) === -1) {
    // It's not in there
}

But if you insist on jQuery:

if ($.inArray($body[0].id, $pages) === -1) {
    // It's not in there
}

answered Jun 13, 2013 at 19:43

Blender

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2 Comments

Blender Over a year ago

@karthikr: OP's question had $body[0].id !== anyOfThePageInTheArray, so it looks like the condition is checking if the element isn't in the array.

Ali Over a year ago

it doesn't work for me I don't know why :/ It is like it never check the whole array and already returned.

karthikr · Accepted Answer · 2013-06-13 19:43:03Z

2

Try:

if($pages.indexOf($body[0].id) == -1){
    //code here
}

answered Jun 13, 2013 at 19:43

karthikr

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Comments

petemir · Accepted Answer · 2016-04-24 00:06:13Z

2

You should use

if($.inArray($body[0].id,$pages) === -1)

When the second argument isn't in the array (first argument), it will return -1. Otherwise, it returns the index of the element.

By the way, $body[0].id returns e.g. "events", or is a number?

edited Apr 24, 2016 at 0:06

answered Jun 13, 2013 at 19:45

petemir

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3 Comments

Ali Over a year ago

I think it is returning the page name for that argument

petemir Over a year ago

Before the if you could debug with console.log($body[0].id); console.log($pages); to know what's going wrong...

Ali Over a year ago

matiaslaporte I think I know what is wrong, I forgot my closing bracket on something, so this code works, but FYI your equal should be ===

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