Loop each jQuery Issue

Question

I have this code:

var arrayInstSaude = new Array();
$("input[name='uniSaudePrj']:checked").each(function(){
    arrayInstSaude[$(this).val()]=$(this).val();
});

For some reason it gives me a messed array. Exemple:

if I check 1 element value eq 1. It gives me the arrayInstSaude length equal 2.
if I check 2 elements value eq 2. It gives me the arrayInstSaude length equal 3.
if I check 3 elements value eq 5. It gives me the arrayInstSaude length equal 6.
if I check 4 elements value eq 6. It gives me the arrayInstSaude length equal 7.
if I check 5 elements value eq 7. It gives me the arrayInstSaude length equal 8.

If I do that for 5 elements:

for (var i = 1; i <=arrayInstSaude.length; i++) {
     alert(arrayInstSaude[i]);
}

I will have 1,2,undefined,undefined,5,6,7,undefined, while it was expected to have 1,2,5,6,7. Someone know what is going on? Thanks!

Array indexes start from 0. If you create array[1], the length is 2 because it has array[0] and array[1]. — Barmar
– Barmar, Commented Nov 4, 2013 at 17:14
When you create arrayInstSaude[5] it automatically creates 3 & 4, but with no values. It can't just leave them out. — Reinstate Monica Cellio
– Reinstate Monica Cellio, Commented Nov 4, 2013 at 17:17

SLaks · Accepted Answer · 2013-11-04 17:13:41Z

3

Arrays are always contiguous.
The length is simply the highest index plus one.

It sounds like you want a regular object that happens to have numeric keys (and no length).

answered Nov 4, 2013 at 17:13

SLaks

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Community · Accepted Answer · 2017-05-23 11:57:32Z

0

Replace

for (var i = 1; i <=arrayInstSaude.length; i++) {
     alert(arrayInstSaude[i]);
}

with

for (var i in arrayInstSaude) {
     alert(arrayInstSaude[i]);
}

Consider reading this discussion by olliej.

edited May 23, 2017 at 11:57

CommunityBot

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answered Nov 4, 2013 at 18:34

Brian

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Comments

ABA · Accepted Answer · 2013-11-05 10:34:33Z

0

I think this code is enough for my propouse:

var arrayInstSaude = new Array();                
var k=0;
$("input[name='uniSaudePrj']:checked").each(function(){                    
    arrayInstSaude[k]=$(this).val();
    k++;
});

It was important to observe that the length is the highest index plus one. So, to retrieve the elements we could use

for(k=0;k<arrayInstSaude.length;k++){
    alert(arrayInstSaude[k]);
}

Because we know that the last element is allways empty.

answered Nov 5, 2013 at 10:34

ABA

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