JavaScript equivalent of PHP $array[$i][]

Question

I have a problem with the creation of a multidimensional array in JavaScript.

The PHP code looks like this:

<?php
$matches = array(1, 2, 3, 4, 4, 4, 6, 6, 2, 3);

foreach($matches as $match) {

     $levels[$match][] = $match;
}

print_r($levels);
?>

print_r of $levels:

levels[1][0] = 1
levels[2][0] = 2
levels[3][0] = 3
levels[4][0] = 4
levels[4][1] = 4
levels[4][2] = 4
levels[6][0] = 6
levels[6][1] = 6
levels[2][1] = 2
levels[3][1] = 3

I have a problem with the creation the same array in JavaScript.

<script>
var levels = [];

$([1, 2, 3, 4, 4, 4, 6, 6, 2, 3]).each(function(key, value) {

     levels[value][] = value;
});
</script>

Can someone help me create the same array in JavaScript?

Answer 1

JavaScript doesn't magically create an array if there is none. You have to create it yourself. So it would be something like

$([1, 2, 3, 4, 4, 4, 6, 6, 2, 3]).each(function(key, value) {
    if (levels[value] == null) {
        levels[value] = [];
    }
    levels[value].push(value);
});

Learn more about arrays.

Answer 2

var levels = [];
var orgArray = [1, 2, 3, 4, 4, 5, 6, 2];
for (var i = 0; i < orgArray.length; i++) {
    var value = orgArray[i];
    if (levels[value] === undefined)
        levels[value] = []
    levels[value].push(value);
}
//Print
for (var j = 0; j < levels.length; j++)
    if (levels[j])
        for (var k = 0; k < levels[j].length; k++)
            console.log("levels[" + j + "][" + k + "] = " + levels[j][k]);

It's a bit overkill but you will get the point I think. Don't forget to check if your elements exists

Answer 3

Felix's answer is good and readable, I just want to introduce you to a couple of useful idioms.

Unlike php, javascript || (boolean OR) operator, despite its name, doesn't return boolean, but rather the first non-falsy operand. So when you write

a = thing || otherThing

the result will be otherThing if thing is falsy (= null, undefined, 0, empty string). Therefore checks like this

if(!foo)
   foo = bar

can be written more concisely as

foo = foo || bar

If foo is truthy, it just assigns foo to foo (a no-op), otherwise, foo becomes bar.

Applied to your problem, this will look like this:

levels[value] = levels[value] || [];
levels[value].push(value);

If you want to shorten this even more, replace push with concat:

levels[value] = (levels[value] || []).concat(value);

This is not necessarily more readable or efficient, just something worth knowing about.