jQuery array parameter and each()

Question

I'm new in jQuery world, still learning. I'm trying to understand how can I pass an array parameter reading data from an "each()" loop. Let my try to explain (my native language is Portuguese, sorry).

My original code is below:

$('#relatorio').DataTable({
   (...)
   columns: [
      {visible:$('#c1').is(':checked')},
      {visible:$('#c2').is(':checked')},
      {visible:$('#c3').is(':checked')}
   ]
});

Since I'll use this in several places (reports), I did a .JS file. But, in each report, I'll have different checkboxes. So I want to scan each checkbox and mount the object array dynamically. I've add a class "colsel" in each checkbox, so I can do:

var vCada=[];
$('.colsel').each(function(){ vCada.push({visible:$(this).is(':checked')}); });

Ok, but now I don't know how to replace this directly in there.

I had success creating a function that returns the object array:

function StatusCols() {
  var vCada=[];
  $('.colsel').each(function(){ vCada.push({visible:$(this).is(':checked')}); });
  return vCada;
}

and then:

columns: StatusCols()

But I'm not satisfied with this and my newbie knowledge is not helping me :(

I've tried:

columns: function() {
   var vCada=[];
   $('.colsel').each(function(){ vCada.push({visible:$(this).is(':checked')}); });
   return vCada;
 }

So, where is my mistake? Can anyone help me (and teach me)?

Thank you!

"But I'm not satisfied with this" Why? The mistake at the end is that you are assigning a function to columns, instead of calling the function and assign its return value. I don't see how the last solution would be any improvement over the original code though. — Felix Kling
– Felix Kling, Commented Mar 18, 2015 at 23:16
You can try this: pastebin.com/raw.php?i=10UX0xpP but columns: StatusCols() is fine. — blex
– blex, Commented Mar 18, 2015 at 23:19
Hi Felix, my previous code works, but need an extra function. I'm trying to remove the additional function, so my idea was create a "local" function that could return an array. — Arvy
– Arvy, Commented Mar 18, 2015 at 23:25
I'm not sure if I've found your exact issue just by looking, but one helpful tip; if you want to make another array that's based on the contents of a first array, it's handy to use .map(. It works similarly to each, but...well, look it up here: api.jquery.com/jQuery.map — Katana314
– Katana314, Commented Mar 18, 2015 at 23:25
@Arvy Let's say you have a function called myFunc. myFunc is equal to function(){...}. You call it by typing its name + a set of parenthesis: myFunc(). Well, it's equivalent to calling it directly (replace the name by the function declaration itself): function(){...}() — blex
– blex, Commented Mar 18, 2015 at 23:28

Arvy · Accepted Answer · 2015-03-18 23:37:10Z

1

Thank you @blex. Solved. Solution:

columns: function() {
   var vCada=[];
   $('.colsel').each(function(){ vCada.push({visible:$(this).is':checked')}); });
   return vCada;
 }()

Means: function(){...}()

answered Mar 18, 2015 at 23:37

Arvy

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