Checking item of array on "undefined" in JavaScript

Question

My JS-code has array arrayResults, some element of him can be "undefined" - this is feature of algorithm. To check that there is no such elements I use the follow code:

for (i in arrayResults)
        {
          if (typeof(arrayResults[i])=='undefined')
           {
              // ask user to repeat
           };
        };

But, using the debugger, I found that JS-engine passes the "undefined"-item of array (in for condition), respectively I don't have the possibility to make the comparing and make the follow instructions.

So, is there any way to really check the "undefined" items in array? (I can't to set items of array in sequence, because if I found the position of "undefined" item, I tell to user to go to this position).

score 9 · Accepted Answer · edited May 23 '17 at 11:54

9

Don't use a for..in loop to iterate arrays. If you are interested in the reasons, please read this StackOverflow question. They should only be used for traversing objects.

Use a simple oldschool for loop instead, it will solve your problem.

for (var i = 0, l = arrayResults.length; i < l; i++) {
    if (typeof(arrayResults[i])=='undefined') {
         // ask user to repeat
    };
};

jsFiddle Demo

edited May 23 '17 at 11:54

Community

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answered Feb 18 '12 at 21:48

kapa

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+1 Here the old-school for loop is superior to all the higher-order functions like `.some()` or `.reduce()` or even `.indexOf()` because the array methods skip so-called "holes", elements that are "undefined" because they were never set. – Paul Mar 15 '17 at 04:08

score 4 · Answer 2 · answered Aug 12 '15 at 07:35

4

You can use indexOf method on array.

function hasUndefined(a) {
    return a.indexOf() !== -1;
}

hasUndefined([1,2,3, undefined, 5]);

answered Aug 12 '15 at 07:35

Mateusz Nowak

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But `hasUndefined([1,2,3,,,,,,,7])` returns `false` so this is not robust. – Paul Mar 14 '17 at 04:04

Checking item of array on "undefined" in JavaScript

2 Answers2