jQuery - Building object arrays with html5 data attribute selection

Question

I'm using HTML 5 data attributes to key rows of data in a table. I need to iterate the rows and gather the data into an object collection. I created a class to represent my row data:

function SomeItem(description, finalPrice, percentDiscount) {
    this.Description = description;
    this.FinalPrice = finalPrice;
    this.PercentDiscount = percentDiscount;
}

An event fires which triggers the collection of this data.

$.each($('.ItemPriceBox'), function () {
            var uniqueid = $(this).data('uniqueid');
            var finalPrice = $(this).val();
        });

The final piece of data, percentDiscount should be retrieved using the data-uniqueid attribute. I'm not sure how to do this.

I want...

SomeItem[] items;
$.each($('.ItemPriceBox'), function () {
            var uniqueid = $(this).data('uniqueid');
            var finalPrice = $(this).val();
            var percentDiscount = $('.ItemDiscountBox').where("uniqueid = " + uniqueid);
            items[i] = new SomeItem(uniqueid,finalPrice,percentDiscount);
        });

How can I solve this?

Rafay · Accepted Answer · 2011-09-13 18:27:26Z

3

may be like this

var percentDiscount = $(".ItemDiscountBox[uniqueid='"+uniqueid +"']");

answered Sep 13, 2011 at 18:27

Rafay

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1 Comment

P.Brian.Mackey Over a year ago

+1 Thats pretty close. I just finished up with a slightly different solution.

Dennis · Accepted Answer · 2011-09-13 18:41:24Z

0

Instead of your pseudo-code where, use filter:

var percentDiscount = $('.ItemDiscountBox').filter("[uniqueid='" + uniqueid + "']");

$('.ItemPriceBox').each( function (i) {
    var uniqueid = $(this).data('uniqueid');
    var finalPrice = $(this).val();
    var percentDiscount = $('.ItemDiscountBox').filter("[uniqueid='" + uniqueid + "']");
    items[i] = new SomeItem(uniqueid,finalPrice,percentDiscount);
});

You can also use .each() instead of jQuery.each()

answered Sep 13, 2011 at 18:41

Dennis

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1 Comment

P.Brian.Mackey Over a year ago

This is the right idea, but 3nigma's solution is more concise.

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