Question

87

jQuery.each() undefined issue

rated 0 times [ 94] [ 7] / answers: 1 / hits: 24460 / 14 Years ago, sun, july 25, 2010, 12:00:00

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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devonw

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answered 14 Years ago tiffany · Accepted Answer

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.