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Home/ Questions/Q 951609
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T23:45:48+00:00 2026-05-15T23:45:48+00:00

I have the following code function myFunction(items) { // this prints out 11 alert(items.length);

  • 0

I have the following code

function myFunction(items) {
   // this prints out 11
   alert(items.length);

   $(items).each(function(i, item) {
       // item is undefined for some reason
   }
}

It I alert the length of items, it has elements in it (11 to be exact). so how could 11 items be present, but jQuery still pass undefined?

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  1. Editorial Team
    Editorial Team
    2026-05-15T23:45:49+00:00Added an answer on May 15, 2026 at 11:45 pm

    The only explanation for this is the that items array contains values which are undefined, i.e :

    items = [undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined];
    

    Both of the other answers are wholly incorrect. The first parameter of each is the index, not the value, and jQuery.fn.each calls jQuery.each. There is no disambiguation between them.

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