Undefined values in Javascript array

Question

I'm trying to add 2 values to an array that I extract from a html page. However, the value for key [0] always returns undefined whereas I expect 4 in this case.

<script type="text/javascript">
 $(document).ready(function ()
 {
 var arrayList = $("some class");
 var x = new Array(); 
 var j = 0;
 var k = 1;
 $.each(arrayList, function(i,e) {
 var MyIf = $(e).text(); 
 x[j] = new Array();
 if(k == 2) {
 x[j][0] = 4; // This always returns undefined, no matter which value I assign.
 }
 if(k == 3) {
 x[j][1] = parseInt(MyIf);
 }
 if(k % 3 == 0) {
 j++;
 k = 1;
 } else {
 k++;
 }
 });
 console.log(x); // the console returns for all [0] "undefined"
 });
</script>

What am I missing?

• Daniël Knippers: How many elements does arrayList have? If it has only 1, I would expect x to be empty since you only add to it when k == 2 or k == 3. In other words, how many times does your $.each loop run?
• Kris Van den Bergh: arrayList has multiple values. Current output looks like this: [[undefined, 3], [undefined, 5], [undefined, 2]]
• Trace: A fiddle will help...
• sdespont: Are you sure that k==2 sometimes?
• mihai: what is this code trying to achieve?

Answer 1

It's because you create a new array every time in your each.

x[j] = new Array();

so it's always undefined.

put it in here:

if(k % 3 == 0) {
 j++;
 k = 1;
 x[j] = new Array();
 }

and don't forget to call x[j] = new Array(); before your first run.

• Kris Van den Bergh: so how does it explain that the Array still has values for x[j][1] ?
• Homungus: because you make a new array an then set the value x[j][1] = somevalue;
• Daniël Knippers: @KrisVandenBergh Because when k == 3 it adds to the last created Array at index 1. Then, j is increased so it will not touch that same Array anymore.
• Daniël Knippers: @Homungus If you do x[j] = [] in the k % 3 == 0 case, x[j] will be undefined in the k == 2 and k == 3 cases above it the very first run. Perhaps not pretty, but you can do on the first line of the .each loop a if(typeof x[j] === 'undefined') x[j] = []; which fixes it (and remove it from the k % 3 == 0 case entirely).
• Homungus: @DaniëlKnippers you are right, of course it should be initialized before first run. updated answer.

Answer 2

It should be done like this:

var arrayList = [1,2,3,4,5];
var x = new Array(); 
var j = 0;
var k = 1;
x[j] = new Array();
$.each(arrayList, function(i,e) {
 var MyIf = $(e).text(); 
 if(k == 2) {
 x[j][0] = 4; // This always returns undefined, no matter which value I assign.
 }
 if(k == 3) {
 x[j][1] = parseInt(MyIf);
 }
 if(k % 3 == 0) {
 j++;
 x[j] = new Array();
 k = 1;
 } else {
 k++;
 }
});
console.log(x);

In your code, when the k variable is equal 3, you overwrite the previous value doing this:

x[j] = new Array();
 if(k == 2) {
 x[j][0] = 4; // This always returns undefined, no matter which value I assign.
 }