jQuery Javascript array 'contains' functionality?

Question

I'm trying to use the jQuery $.inArray function to iterate through an array and if there's an element whose text contains a particular keyword, remove that element. $.inArray is only returning the array index though if the element's text is equal to the keyword.

For example given the following array named 'tokens':

-   tokens  {...}   Object
    [0] "Starbucks^25^http://somelink"  String
    [1] "McDonalds^34^" String
    [2] "BurgerKing^31^https://www.somewhere.com"   String

And a call to removeElement(tokens, 'McDonalds'); would return the following array:

 -  tokens  {...}   Object
    [0] "Starbucks^25^http://somelink"  String
    [1] "BurgerKing^31^https://www.somewhere.com"   String

I'm guessing this may be possible using the jQuery $.grep or $.each function, or maybe regex. However, I'm not familiar enough with jQuery to accomplish this.

Any help would be appreciated!

outis · Accepted Answer · 2010-04-06 22:21:26Z

4

grep is indeed the way to go.

function removeElement(array, keyword) {
    return $.grep(array, function (item, i) {
      return item.indexOf(keyword) > -1;
  }, true);
}

This looks for the keyword as a substring. If the elements of the array have a particular format (which they appear to, but it isn't stated for certain), alter the test to match the format.

function removeElement(array, keyword) {
    var keyRE=new RegExp('^'+keyword+'^');
    return $.grep(array, function (item, i) {
      return keyRE.test(item);
  }, true);
}

Note the first '^' is a meta-character matching the beginning of the string; the second simply matches a '^' character.

edited Apr 6, 2010 at 22:21

answered Apr 6, 2010 at 22:13

outis

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3 Comments

Pointy Over a year ago

I think he wants the result array to contain only things that don't have the keyword.

YourMomzThaBomb Over a year ago

Pointy has it right. Although if the first removeElement function you've created returned the array index instead of the element, that would work too?

outis Over a year ago

@YourMomzThaBomb: in the first version (actually, both), the array index is ignored. a.indexOf(b) returns the starting index of string b in string a, or -1 if b isn't a substring of a.

jitter · Accepted Answer · 2010-04-06 22:24:25Z

0

This should be clear on its own. Removes all elements from the tokens parameter which contain the string passed in in the search parameter. Using filter()

function removeElement(tokens, search) {
    var regex = new RegExp(search);
    return $(tokens).filter(function() {
        return !regex.test(this);
    }).toArray();
}

If you want you can also specify flags for the regex

e.g.

var regex = new RegExp(search, "i");

Would remove all elements which contain the search string with case ignored

$.grep version of the code

function removeElement(tokens, search) {
    var regex = new RegExp(search);
    return $.grep(tokens, function(elem, ind) {
        return regex.test(elem);
    }, true);
}

edited Apr 6, 2010 at 22:24

answered Apr 6, 2010 at 22:13

jitter

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1 Comment

Pointy Over a year ago

Building a new RegExp instance for each iteration seems kind-of wasteful.

Pointy · Accepted Answer · 2010-04-06 22:55:31Z

0

edit oops I totally misunderstood the question.

Well $.grep takes an "invert" flag, so you can use it directly:

var pattern = /\bword\b/;
var newArray = $.grep(oldArray, function(e) { return pattern.test(e); }, true);

Of course, now that I think about it, I could save 5 characters:

var newArray = $.grep(oldArray, function(e) { return !pattern.test(e); });

edited Apr 6, 2010 at 22:55

answered Apr 6, 2010 at 22:12

Pointy

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