How do I create a Multidimensional Array in Javascript?

Question

My goal is to iterate over a table with an ID of ajaxResponse1 and store each COLUMN'S values in a separate array. Each of the entries in valuesArr will be a column containing an array of individual values.

I tried using new Array instead of new Object, but I still don't seem to be getting any output alerted to my screen. Should I be using some combination of .push to get each new column value into the appropriate array? Hope that's clear. Any ideas? Thanks!

var valuesArr = new Object();

$('#ajaxResponse1 tr').each( function() {
    var colNum = 0;
    $('th, td', this).each( function() {
        valuesArr[colNum] = valuesArr[colNum].push( $(this).html() );   
        colNum++;
    });
});

alert(JSON.stringify(valuesArr));

Try putting your alert inside the loop so you can check what's going into the array — mafue
– mafue, Commented May 4, 2012 at 16:51
not a duplicate IMHO, since this has the added complication of creating the transpose of the original table. — Alnitak
– Alnitak, Commented May 4, 2012 at 17:13

Alnitak · Accepted Answer · 2012-05-04 17:03:52Z

2

You can't push something onto an array until that array actually exists.

So in each iteration of your inner loop, if the initial array doesn't exist you have to create it:

var valuesArr = [];

$('#ajaxResponse1 tr').each( function() {
    $('th, td', this).each( function(i, el) { // nb: implicit column number
        var html = $(this).html();
        if (i in valuesArr) {
            valuesArr[i].push(html);
        } else {
            valuesArr[i] = [ html ];  // new array with one element
        }
    }
});

edited May 4, 2012 at 17:03

answered May 4, 2012 at 16:54

Alnitak

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2 Comments

Andrew Over a year ago

Worked like a charm. I NEVER would have guessed that was the issue. Thanks for the help!

Alnitak Over a year ago

@Anjroo see updated code example which eliminates the colNum variable

Samuel Hapak · Accepted Answer · 2012-05-04 17:03:35Z

1

The push() method adds new items to the end of an array, and returns the new length. So, try to change it:

valuesArr[colNum].push( $(this).html() );

Just remove the =

Moreover, you need to initialize array before first time using it, try:

valuesArr[colNum] = valuesArr[colNum] ? valuesArr[colNum] : [];

You will find complete working example here http://jsfiddle.net/57A9z/

edited May 4, 2012 at 17:03

answered May 4, 2012 at 16:52

Samuel Hapak

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Comments

mihai · Accepted Answer · 2012-05-04 16:59:53Z

0

var valuesArr = [];
var r = 0;
var c = 0;

$('#ajaxResponse1 tr').each( function() {
    valuesArr[r] = [];
    $('th, td', this).each( function() {
        valuesArr[r][c] = $(this).html();   
        c++;
    });
    c = 0;
    r++;
});

edited May 4, 2012 at 16:59

answered May 4, 2012 at 16:52

mihai

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2 Comments

Alnitak Over a year ago

You're creating an entry for each row, not one for each column.

mihai Over a year ago

yep, I misunderstood the question. Looking at the code I thought it made more sense this way, but it seems like the requirements were different. cheers :)

nem0z · Accepted Answer · 2022-07-21 13:51:23Z

0

Multidimensional array = embeded arrays. See if this tiny model can help. thanks.

   "use strict";
   const arr = [
       ["D1","D2","D3"],
       [
           ["T11","T12","T13"],
           ["T21","T22","T23"]
       ]
   ];

   for(let k=0;k<arr[0].length;k++)
      console.log(arr[0][k]);
      // D1
      // D2
      // D3

   for(let k=0;k<arr[1].length;k++)
      console.log(arr[1][k]);
      // Array(3) [ "T11", "T12", "T13" ]
      // Array(3) [ "T21", "T22", "T23" ]

   for(let k=0;k<arr[1].length;k++)
      console.log(arr[1][0][k]);
      // T11
      // T12

   for(let k=0;k<arr[1].length;k++)
      console.log(arr[1][1][k]);
      // T21
      // T22

edited Jul 21, 2022 at 13:51

nem0z

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answered Jul 20, 2022 at 11:41

Doct Jean-Dadet Luyalu

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