jquery Datatables filter rows

Question

Hi I have a DataTables DataTable object which requests json data via an ajax call. Each object inside the json data has a property called state that can be one of a number of values. Ultimately, I'd like to create several (Data)tables, one for each state without having each table request the same data via ajax again and again. For now, I haven't managed to make a single table filter out the rows with incorrect state yet and I'd like to solve this problem first.

The DataTable object defined as follows:

$(document).ready(function () {
    var table = $('#example').DataTable({
        data: example,
        ajax: {
            url: "{{ callback_url }}",
            dataType: 'json',
            dataSrc: '',
        },
        createdRow: function (row, data, index) {
        },
        columns: [{
            data: "Type"
        }, {
            data: "State",
        }
        }]
   });
});

I want to filter the data that comes from the ajax call based on a parameter (e.g. "if (row.State == 'new') {...};"). How can I do that? Does datatables have a function that can be passed to filter each row?

I really need several tables which show data from the same ajax callback but in different states (e.g. a "New" table, an "Old" table etc.). How can I do this without requesting the same json data again and again for each table rendered? Ideally I can store the ajax in a variable but with the async process I'm not sure how this will work out.

weigreen · Accepted Answer · 2016-06-07 13:03:59Z

2

You can use one ajax and store the data, and than create each datatable.

Example:

$(document).ready(function () {
  var data = null;
  $.ajax(URL, {
    dataType: 'json',
    success: function(ajaxData){
        data = ajaxData;

      newData = data.filter(function(item){ return item.State == "New"});
      oldData = data.filter(function(item){ return item.State == "Old"});

      var newTable = $('#newTable').DataTable({
        data: newData,
        });

      var oldTable = $('#oldTable').DataTable({
        data: newData,
        });

    }
  });
}

answered Jun 7, 2016 at 13:03

weigreen

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