Why does this simple javascript array fail?

Question

I am getting old or just rusty or both, but I hang my head in shame as I bring this here because there must be something really, really simple that I am missing.

I am writing this using Google Apps script in Sheets.

Why does this fail once I reach j = 2? I have tried declaring the array in every different way I can think of, nothing gets past j=2. Wth am I missing? It's something dumb I know it.

function myFailure() {
  for (var j = 0; j < 10; j++) {
    for (var k = 0; k < 31; k++) {
      var item = 'Item '+k;
      let thisItem = new Array([],[]);
      thisItem[j][k] = item;  //the problem is happening here, once j=2 but why
      console.log(j,k);
      console.log(thisItem[j][k]);
    }
  }
}

myFailure();

console.log(thisItem) and I think this will point you out to the solution ;) — miyav miyav
– miyav miyav, Commented Apr 12, 2022 at 15:35
new Array([], []) creates an array with 2 arrays in it. When j is 0 or 1, it works, but not when it's 2. — Pointy
– Pointy, Commented Apr 12, 2022 at 15:35
In general it's better to use the myArray = [] format instead of new Array(). Check out this answer as a reference. — Daniel
– Daniel, Commented Apr 12, 2022 at 17:13

Cooper · Accepted Answer · 2022-04-12 16:03:29Z

2

Try this:

function myFailure() {
  let thisItem = [];
  for (var j = 0; j < 10; j++) {
    thisItem[j] = []
    for (var k = 0; k < 31; k++) {
      thisItem[j][k] = `Item [${j}][${k}]`
      console.log(thisItem[j][k]);
    }
  }
}

Partial Execution log

10:04:00 AM Info    Item [0][0]
10:04:00 AM Info    Item [0][1]
10:04:00 AM Info    Item [0][2]
10:04:00 AM Info    Item [0][3]
10:04:00 AM Info    Item [0][4]
10:04:00 AM Info    Item [0][5]

answered Apr 12, 2022 at 16:03

Cooper

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1 Comment

Dave Over a year ago

Thank you! I see the issue now. I was bunching up my array in the loop.

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