Solved

jQuery $.each not working for array

Posted on 2014-02-05
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463 Views
Last Modified: 2014-02-06
Hi,

I have this bit of js code that grabs all the keys and values for a form:

    $.fn.calculate_selections.get_all_selections = function() {
        var inputData = new Array();

        $("#get_all_selections").find('input[type="text"]').each(
            function (unusedIndex, child) {
                inputData[child.name] = child.value;
            });

        $("#get_all_selections").find('input[type="hidden"]').each(
            function (unusedIndex, child) {
                inputData[child.name] = child.value;
            });

        $("#get_all_selections").find('input[type="radio"]:checked').each(
            function (unusedIndex, child) {
                inputData[child.name] = child.value;
            });

        $("#get_all_selections").find('select option:selected').each(
            function (unusedIndex, child) {
                var oClosest = $(this).closest('select');
                var sName = oClosest.attr('name');
                inputData[sName] = child.value;
            });

        $("#get_all_selections").find('textarea').each(
            function (unusedIndex, child) {
                inputData[child.name] = child.value;
            });
        return inputData;

    }

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I return it from a plugin:
var aInputDataVals = $('').calculate_selections.get_all_selections();

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But this part doesn't work:
    $.each(aInputDataVals, function (key, value) {
        calculate_final_amount(key, value);
    });

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What am I doing wrong? I'm trying to debug in Chrome and it's not stepping into the $.each(). I can see aInputDataVals is an array of keys and values.

aInputDataVals['test'] = 1;
0
Comment
Question by:Victor Kimura
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2 Comments
 
LVL 22

Accepted Solution

by:
Mrunal earned 500 total points
ID: 39837825
hi,
There may be issue with your creating array logic.

Please find below code to repeat among array items and create/update logic for creating array:

$(document).ready(function(){
    var obj = { one: 1, two: 2, three: 3, four: 4, five: 5 };    
    var arr = [ "one", "two", "three", "four", "five" ];
    alert("obj items are: ");
    $.each(obj, function(index, value) {
        alert(index + ": " + value);
    });
    
    alert("arr items are: ");
    $.each(arr, function(index, value) {
        alert(index + ": " + value);
    });
});

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hope this helps you.
0
 

Author Comment

by:Victor Kimura
ID: 39839409
Hi Mrunai,

Yes, if I use the array like in your example var arr then I found the indexes are 0, 1, 2, etc.

but the function I have must use var obj = {} like in your example since it's actually an object not an array that is created I think. Anyways, changing the line from:

var inputData = new Array();

to:

var inputData = {};

does the trick.

Thank you! =)
0

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