How to check if string is .inarray

Question

How can I check if a string is contained with my array? Here's how I'm putting the array together...

 // get the select
 var $dd = $('#product-variants');

 if ($dd.length > 0) { // make sure we found the select we were looking for

 // save the selected value
 var selectedVal = $dd.val();

 // get the options and loop through them
 var $options = $('option', $dd);
 var arrVals = [];
 $options.each(function(){
     // push each option value and text into an array
     arrVals.push({
         val: $(this).val(),
         text: $(this).text()
     });
 });
};

I want to check if "Kelly Green" is contained within the array if it is I want to .show a li

$("#select-by-color-list li#kelly-green").show();

Nick Craver · Accepted Answer · 2010-12-31 11:29:16Z

4

The other answers are correct so far as use $.inArray() here, however the usage is off, it should be:

if($.inArray('Kelly Green', arrVals) != -1) {
  $("#select-by-color-list li#kelly-green").show();
}

$.inArray() returns the position in the array (which may be 0 if it's first...so it's in there, but that if() would be false). To check if it's present, use != -1, which is what it'll return if the element is not found.

answered Dec 31, 2010 at 11:29

Nick Craver

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Comments

Tatu Ulmanen · Accepted Answer · 2011-01-02 14:55:14Z

1

You can use jQuery's inbuilt .inArray() method:

if($.inArray('Kelly Green', arrVals) != -1) {
    $("#select-by-color-list li#kelly-green").show();
}

edited Jan 2, 2011 at 14:55

answered Dec 29, 2010 at 16:17

Tatu Ulmanen

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2 Comments

Nick Craver Over a year ago

Be careful using this, the check is incorrect, it should be if($.inArray('Kelly Green', arrVals) != -1) {, 0 is a valid result, if the element is the first in the array you'll get 0, -1 is what you should be checking for.

Tatu Ulmanen Over a year ago

@Nick Craver, you're right, I've fixed the answer. I naturally assumed that jQuery would not duplicate JS's own idiocies and would return rather just a boolean value indicating the precense of a value. :)

Community · Accepted Answer · 2017-05-23 12:26:56Z

1

Try this:

console.log($.inArray('Kelly Green', arrVals);

if($.inArray('Kelly Green', arrVals)
{
  $("#select-by-color-list li#kelly-green").show();
}

Possible dupe: Need help regarding jQuery $.inArray()

edited May 23, 2017 at 12:26

CommunityBot

11 silver badge

answered Dec 29, 2010 at 16:18

user67416

2 Comments

Tatu Ulmanen Over a year ago

You got the parameters wrong way around, it's .inArray(value, array).

Charles Marsh Over a year ago

@ Noah and Tatu, Nice one! I was thinking about it to hard! Works great! Thanks very much

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