JavaScript
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jQuery selector first x items of specific class

By Sam Deering

I’ve been playing around with this a little. Grabbing a specific number of DOM elements based on a specified class.

Update 27/11/12: Optimised versions below.

First, basics to get the first and last elements like so:

var firstSpan = $('span.class:first'),
    lastSpan = $('span.class:last');

To get all the elements that match a specified class like so:

var allSpans = $('span.class').get();

or the nth/xth element like so:

var firstSpan = $('span.class').get(0),
    secondSpan = $('span.class').get(1);
    //etc...

But how to get say the first 10 elements or elements 10-20?

It would be nice to go something like:

var mySpans = $('span.class').get(0,10);

Unfortunately, the .get() function doesn’t allow for a range to be passed, but just a single index. So here is an attempt to use the jQuery .get() function to include a range of elements.

(function($)
{
  //function that gets a range of dom elements against a jQuery selector
  //returns an array of dom elements
  $.fn.getRange = function(start,end)
  {
    var elems = [];
    for ( var i = start; i < = end; i++ )
    {
      elems.push(this.get(i));
    }
    return elems;
  };

  //testing
  console.log($('div').getRange(1,10));
  console.log($('div').getRange(10,20));

})(jQuery);
[/js]

Does anyone know of a better way to achieve this?

Optimised versions Thanks to Vlad, Redky and Daniel. Using slice. [js] var $el = $('div').get().slice(0,10); console.log($el);​

Using :gt :lt

console.log($('div:gt(3):lt(6)'));​
  • http://www.facebook.com/vladimir.kharlampidi Vladimir Kharlampidi

    Yes, there is a more better native way to achieve this using :gt() and :lt() pseudo selectors and their combinations like:
    $(‘div:gt(5):lt(10)’)

    • http://jquery4u.com/ jQuery4u

      Hi Vlad, doesn’t seem to work mate, am I doing something wrong? http://jsfiddle.net/YHprN/
      Sam

      • Daniel Trott

        I tested it a little bit and came to the following conclusion:
        With the given div-set from your fiddle and using :gt(5) you first get all divs from index 6, because :gt() and lt() are zero-based indices. From this set you try to get all divs less than the index 10 ( :lt(10) ).
        You can try to set the :lt() index to 1 and you’ll get just one div. So the conclusion here is, that you get first all divs greater than index 5 (div#7 – div#15) and then all divs within this set until index less than 10 (still div#7 – div#15, because there are less then 11 elements).
        On the other hand the slice-operation works well.

        • http://www.facebook.com/vladimir.kharlampidi Vladimir Kharlampidi

          That is right. To output divs from #6 to #10, you need to type $(‘div:gt(4):lt(5)’)

          • http://jquery4u.com/ jQuery4u

            Yes this would give you els 5-10 $(‘div:gt(3):lt(6)’) – http://jsfiddle.net/YHprN/2/
            My conclusion is with minimum overhead increase by getting all elements initially but slice works best.
            Sam

  • redky

    We also can use Array.prototype.slice method to achieve this

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