Use jquery function .on

Question

How to use .on on load event more then one time?

var $CartContent = function(){
    $Handler = $('#cartHandler');
    if($Handler.length > 0){
        $Handler.load('cart/content', function(){
            $PercentField = $('#item-percent-field');
            $PercentField.val(0);
            $('.percent-field').val(0);
            $PercentField.on('keyup', function(){
                var $Value = parseInt($(this).val());
                $.each($('.percent-field'), function(index, value){
                    var $Current = $(this).data('item');
                    $.post('cart/content', {
                        id: $Current,
                        percent: $Value
                    });
                });
            });
        });
    }
};

It work's fine only one time. How to start function load with parameters on every keyup?

Denys Séguret · Accepted Answer · 2012-10-22 08:09:12Z

2

You may use the keyup function :

$('#item-percent-field').keyup(function(){
    $('#cartHandler').load( ...
    ...
});

But be sure to have a fast server if you want to load a page on every key up.

Regarding the whole problem, it's not easy to tell what you're trying to do.

First a few remarks :

do you really want to have a function named $CartContent ? Why ?
you don't need to test if($Handler.length > 0){ : if empty the following line would simply do nothing
do you want to make a post request for every percent-field each time there is one changed ? This is strange

edited Oct 22, 2012 at 8:09

answered Oct 22, 2012 at 7:57

Denys Séguret

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5 Comments

Denys Séguret Over a year ago

Nothing really wrong but when you have a selector with id, and which isn't dynamically added, there is no reason to take the long (and slow) path.

RDK Over a year ago

Аfter $PercentField.on('keyup', function(){}); I need to update results with .load $Handler.load('cart/content') after this my code no working

Denys Séguret Over a year ago

What you want to do is unclear for me. What's the logic of those requests ?

RDK Over a year ago

What's the logic of those requests ? Update Session Data, and get data

Denys Séguret Over a year ago

But why do you make, each time one field is changed, one post request per field ? Is this a feature or a bug ?

Samuel Cook · Accepted Answer · 2012-10-22 08:03:12Z

I would just move the $PercentField function out altogether and switch to the .live() method, since #item-percent-field won't be created until after the .load is complete

$(function(){
    $('#item-percent-field').live('keyup',function(){
        var $Value = parseInt($(this).val());
        $.each($('.percent-field'), function(index, value){
            var $Current = $(this).data('item');
            $.post('cart/content', {
                id: $Current,
                percent: $Value
            });
        });
    })
    var $CartContent = function(){
        $Handler = $('#cartHandler');
        if($Handler.length > 0){
            $Handler.load('cart/content', function(){
                $PercentField = $('#item-percent-field');
                $PercentField.val(0);
                $('.percent-field').val(0);;
            });
        }
    });
});

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