CODEWITHSUNDEEP
jqueryfilter
Phill Pafford
Phill Pafford

Reputation: 85308

jQuery filter dropdown

Ok I have a drop down of countries and only want to display USA and Canada

I'm looking at something like this but don't think I'm doing this right

// Filter Countries
$('#selectId').filter(function() {
   return $('#option').val() == ('USA' || 'CANADA');   
});

Example HTML:

<select id="country">
  <option value="AFG">Afghanistan</option>
  <option value="ALB">Albania</option>
  <option value="ALG">Algeria</option>
  <option value="CANADA">Canada</option>
  <option value="USA">United States</option>
</select>

Upvotes: 1

Views: 3860

Answers (3)

Patrick McElhaney
Patrick McElhaney

Reputation: 59271

The || construct you have is invalid. Try this.

$(this).val() == 'USA' || $(this).val() == 'CANADA';

To remove the options that don't match, you could do something like this.

$().ready( function () {
  $('#country option').filter(function() {             
     return !($(this).val() == 'USA' || $(this).val() == 'CANADA');   
  }).remove();
});

Upvotes: 8

Gabriel Hurley
Gabriel Hurley

Reputation: 40052

There's several problems there... you need to use the proper selector #country, filter only returns objects, it doesn't actually act on them... the compound OR syntax there doesn't work in JS...

This works (tested):

$('#country').children().each(function() {
   if ($(this).val() != 'USA' && $(this).val() != 'CANADA') $(this).remove();   
});

You have to use the remove() method because hide() won't do the trick for option elements.

Upvotes: 1

cgp
cgp

Reputation: 41381

Alternatively, if you have a lot of items in your list you want to filter against, or you want to change the items you are filtering for, use this:

// define this anywhere
Array.prototype.inArray = function (value) {
    var i;
    for (i=0; i < this.length; i++) {
        if (this[i] === value) {
            return true;
        }
    }
    return false;
};

// your choice on how to provide this list, I'm using
// something simple as an example
var want = ['USA', 'CANADA'];

// now, somewhere, this filter happens
$('#selectId').filter(function() {
   return want.inArray($('#option').val());   
});

Upvotes: 1

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