yadcf Is it possible to give each option a separate data soruce? And refresh?

Question

DataTable initialisation, give it the server url location

oTable = $('.entrys_table').DataTable({
     "processing": true,
     "serverSide": true,
     "ajax": {
          "url": "entrys_table_server_side_source",
          "type": "POST"
     },
     "language": {
          "infoFiltered": " - filtered from _MAX_ records"
     },
     "columns": [
          { data: "engine" },
          { data: "browser" },
     ]
})

yadcf.init(oTable, [
     {
          column_number: 0,
          filter_container_id: 'account1',
          filter_type: 'select',
     },
     {
          column_number: 1,
          filter_container_id: 'account2',
          filter_type: 'select',
     },
], { externally_triggered: true });

But I would like to ask how to make yadcf URL location is different?

More hope is that each option can have a separate URL location?

And let the option can be updated independently of the data (again to the server to get the latest information)

Any help is appreciated.

score 1 · Accepted Answer · answered Feb 06 '18 at 07:44

The only possible way that can I think of is to use yadcf along with third party select plugin (chosen / select2) and load its data using ajax

Something like Jquery Chosen plugin - dynamically populate list by Ajax or Select2 Ajax (remote data)

In addition please take a look at the following issue that was eventually resolved and looks like it might be useful for you ajax populated fields + externally_triggered + server_side removing fields value

yadcf Is it possible to give each option a separate data soruce? And refresh?

1 Answers1