How to build an Array of Array with JavaScript?

Question

My output should be like this: (eg. data)

myArray = [ [otherArray0], [otherArray1], [otherArrayN] ]

My try:

var myArray = [];
var num = 50;
for (i=0; i<num;i++)
{

    var a = 0;
    $('#divID .class select[value!=""]').each(function() //get values from a select
    {
        var otherArray+i = []; //each for cycle is a new "sub array"
        otherArray+i[a] = $(this).val(); //store in array the values from select
        a++
     })
}

Am I thinking well?

var myArray[i] = array (); should be var myArray[i] = []; — Adam Rackis
– Adam Rackis, Commented Jan 31, 2013 at 17:07
It's really unclear what you're trying to get. You don't put the names of arrays in other arrays, you put the arrays themselves. What do you want the resulting data to look like? — Barmar
– Barmar, Commented Jan 31, 2013 at 17:19
@Barmar, The output should be: myArray[[otherArray1],[otherArray2],[otherArrayN]]where each otherArray has diferent values stored. — noneJavaScript
– noneJavaScript, Commented Jan 31, 2013 at 17:30
That's not possible output. Do you mean [["otherArray1"],["otherArray2"], ...]? — Barmar
– Barmar, Commented Jan 31, 2013 at 17:32

Jamiec · Accepted Answer · 2013-01-31 17:08:02Z

7

Firstly

otherArray[i] = []; // create a new array

Then

otherArray[i][a] = $(this).val();

But your code can be made a lot simpler

var myArray = [];
var num = 50;
for (i=0; i<num;i++)
{
    myArray.push($.map($('#divID .class select[value!=""]'),function(e){
        return e.val();
    }));
}

answered Jan 31, 2013 at 17:08

Jamiec

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3 Comments

noneJavaScript Over a year ago

could it be otherArray+i[i][a] = $(this).val(); ? That +i its just to create diferent arrays

Jamiec Over a year ago

@skdnewbie = otherArray+i makes no sense and is a syntax error as you cant add an integer and Array

noneJavaScript Over a year ago

look at this: otherArray1[1,2,3,4], otherArray2[6,7,8,9] and final should be myArray[[otherArray1],[otherArray2]]

Plynx · Accepted Answer · 2013-01-31 17:23:34Z

2

A pattern of setting up an empty array ([]) then using push to fill it as you go, then pushing the array onto your final array should take you far.

var myArray = [];
for (int i=0; i<50; i++)
{
    var otherArray = [];
    $(/*...*/).each(function()
    {
       var val = /*...*/;
       otherArray.push(val);
    });
    myArray.push(otherArray);
}

edited Jan 31, 2013 at 17:23

answered Jan 31, 2013 at 17:12

Plynx

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3 Comments

noneJavaScript Over a year ago

Hum, Ill try this! Looks like what I want

noneJavaScript Over a year ago

The output should be: myArray[[otherArray1],[otherArray2],[otherArrayN]]where each otherArray has diferent values. Your code will do it ?

Plynx Over a year ago

@skdnewbie Yes. It produces an array of arrays, i.e. [ [], [], [], [] ]

Barmar · Accepted Answer · 2013-01-31 17:43:49Z

1

Reading between lots of lines, I think this is what you want:

var myArray = [];
$("#divID .class select").each(function() {
    var subarray = [];
    $("option[value!='']", $(this)).each(function() {
        subarray.push($(this).val());
    });
    myArray.push(subarray);
}

myArray will then contain:

[[sel1Opt1, sel1Opt2, sel1Opt3, ...], [sel2Opt1, sel2Opt2, sel3Opt3, ...], ...]

answered Jan 31, 2013 at 17:43

Barmar

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Comments

luckystars · Accepted Answer · 2013-01-31 17:12:59Z

0

var num = 50;
for (i=0; i<num;i++)
{
var myArray[i] = array ();
var a = 0;
$('#divID .class select[value!=""]').each(function() //get values from a select
{
    myArray[i].push( $(this).val()); //store in array the values from select

})
}

answered Jan 31, 2013 at 17:12

luckystars

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