How to use a result outside the variable in which it was defined

Asked Jan 17 '20 at 20:18

Active Jan 17 '20 at 20:55

Viewed 32 times

I am defining a DataTable like this:

var yadcf_data_2;
var oTable = $('#example').DataTable({
  "serverSide": true,
  "ajax": {"url": "/platform/elements/?format=datatables",
           "dataSrc": function(json){
             yadcf_data_2 = json.options.yadcf_data_2
             console.log(yadcf_data_2)
             return json.data;
            }
          }

How would I use yadcf_data_2 outside of the variable in which it was defined, like this?

yadcf.exRefreshColumnFilterWithDataProp(oTable, 3, yadcf_data_2);

edited Jan 17 '20 at 20:53

asked Jan 17 '20 at 20:18

OverflowingTheGlass

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You'll need to include it as part of your return, OR, define the variable at the scope where you need it – devlin carnate Jan 17 '20 at 20:31
could you please provide an example? – OverflowingTheGlass Jan 17 '20 at 20:34
`var yadcf_data_2; var oTable = ...` (so you're declaring the variable outside of the scope of the ajax call). – devlin carnate Jan 17 '20 at 20:49
I actually have that already (just added to the question), but a `console.log(yadcf_data_2)` outside of `var oTable = ` results in `undefined` – OverflowingTheGlass Jan 17 '20 at 20:54
That means a value hasn't yet been assigned...or was assigned as undefined – devlin carnate Jan 17 '20 at 20:57
but `console.log(yadcf_data2` within the `oTable` declaration prints the list of values appropriately – OverflowingTheGlass Jan 17 '20 at 20:58
Timing is everything.... – devlin carnate Jan 17 '20 at 21:05
I know - that's why I'm trying to understand how to reference yadcf_data_2 outside of the variable appropriately. I understand it's because of the asynchronous nature, but I'm having trouble applying the answer in the other question to my code – OverflowingTheGlass Jan 17 '20 at 21:08

How to use a result outside the variable in which it was defined

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