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How to display the column headers dynamically in jquery data table

I have the below code for displaying array of objects having property and a value in a data table. But here the column headers are hardcoded as seen in my below html code. How can I make it dynamic based on the input dataset?

 var dataSet = [{
  "Latitude": 18.00,
  "Longitude": 23.00,
  "Name": "Pune"
}, {
  "Latitude": 14.00,
  "Longitude": 24.00,
  "Name": "Mumbai"
}, {
  "Latitude": 34.004654,
  "Longitude": -4.005465,
  "Name": "Delhi"
},{
  "Latitude": 23.004564,
  "Longitude": 23.007897,
  "Name": "Jaipur"
}];
$(document).ready(function() {
    $('#example').DataTable( {
        data: dataSet,
        "columns": [
            { "data": "Name" ,"title":"Custom Name"},
            { "data": "Latitude" },
            { "data": "Longitude" },

        ]
    } );
} );




<table id="example" class="display" cellspacing="0" width="100%">
        <thead>
            <tr>
                <th>Name</th>
                <th>Latitude</th>
                <th>Longitude</th>

            </tr>
        </thead>

    </table>
like image 495
Bruce Avatar asked Dec 11 '22 14:12

Bruce


2 Answers

Assuming the structure of the objects in the dataSet does not change, you could use the first object to build the json object external to the DataTable declaration. If the objects are not of a consistent structure, then you can tweak the logic inside the $.each structure to handle that.

Here's a quick hack:

var dataSet = [{
  "Latitude": 18.00,
  "Longitude": 23.00,
  "Name": "Pune"
}, {
  "Latitude": 14.00,
  "Longitude": 24.00,
  "Name": "Mumbai"
}, {
  "Latitude": 34.004654,
  "Longitude": -4.005465,
  "Name": "Delhi"
}, {
  "Latitude": 23.004564,
  "Longitude": 23.007897,
  "Name": "Jaipur"
}];

var my_columns = [];

$.each( dataSet[0], function( key, value ) {
        var my_item = {};
        my_item.data = key;
        my_item.title = key;
        my_columns.push(my_item);
});

$(document).ready(function() {
  $('#example').DataTable({
    data: dataSet,
    "columns": my_columns
  });
});

You should also consider removing all the static table content in your HTML like this

<table id="example" class="display" cellspacing="0" width="100%"></table>

Here's the jsFiddle https://jsfiddle.net/z4t1po8o/18/

Have fun.

like image 50
nanytech Avatar answered Jul 08 '23 20:07

nanytech


Here is the answer to fetch it from external json

HTML

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
  <script type="text/javascript" charset="utf8" src="https://cdn.datatables.net/1.10.16/js/jquery.dataTables.js"></script>
<link rel="stylesheet" type="text/css" href="https://cdn.datatables.net/1.10.16/css/jquery.dataTables.css">
<script type="text/javascript" charset="utf8" src="script.js"></script>
<div><table id="example"></div>

script.js

jQuery( document ).ready( function( $ ) {
        // Code using $ as usual goes here.

        $.ajax( {
            url: '1.json',
            dataType: "json",
            "success": function ( json ) {
              var tables = $("#example").DataTable({               
                dom : 'l<"#add">frtip',
                "language": {
    "paginate": {
      "previous": "<&nbsp;Previous",
      "next": "next&nbsp;>"
    }
  },
                "columnDefs": [ {
    "targets": 2,
    "createdCell": function (td, cellData, rowData, row, col) {
      if (( cellData > 3 ) && ( cellData < 30 )){
        $(td).css('color', 'green')
      }
      else
           if (( cellData > 31 ) && ( cellData < 50 )){
        $(td).css('color', 'orange')
      }
      else
           if (( cellData > 51 ) && ( cellData < 100 )){
        $(td).css('color', 'red')
      }
    }
  } ],
                "ajax": {
                    "url": "1.json",
                    "type": "POST",
                    "datatype": "json"
                },
                "columns": json.columns
            });
            $('<label/>').text('Search Column:').appendTo('#add')
$select = $('<select/>').appendTo('#add')
$('<option/>').val('0').text('Index').appendTo($select);
$('<option/>').val('1').text('Name').appendTo($select);
$('<option/>').val('2').text('Age').appendTo($select);
$('<option/>').val('2').text('Image').appendTo($select);
$('input[type="search"]').on( 'keyup change recheck', function () {
  var searchValue = $(this).val();
  var columnSearch = $select.val();

  $('#example').DataTable().columns().every(function() {
    //alert('hi');
        this.search('');
    }); 

  if(columnSearch == 'All'){
    tables.search(searchValue).draw();
  } else {
    tables.columns(columnSearch).search(searchValue).draw();
  }
 });
 $select.on('change', function() {
 $('#example').DataTable().search('').draw();
 $('input[type="search"]').trigger('recheck');
 });
            },

        } );
    });

1.json

{
   "data":[
      {
         "Index": 4,
         "Name": "Bob",
         "Age": 7,
         "Image": "None"
      },
      {
         "Index": 2,
         "Name": "Timmy",
         "Age": 4,
         "Image": "None"
      },
      {
         "Index": 3,
         "Name": "Heather",
         "Age": 55,
         "Image": "ddd"
      },
      {
         "Index": 5,
         "Name": "Sally",
         "Age": 22,
         "Image": "None"
      }
   ],
    "columns": [
        { "title": "Index", "data" : "Index" },
        { "title": "Name",  "data": "Name" },
        { "title": "Age", "data": "Age" },
        { "title": "Image", "data": "Image" }
    ]
}
like image 20
ankush Avatar answered Jul 08 '23 20:07

ankush