Answers

• kthorngren Posts: 21,351Questions: 26Answers: 4,955

  July 2018 edited July 2018 Answer ✓

  Are you using the columns.data option to define your columns?

  If not please read the Data Source Types section of the Data manual to see how to use columns.data.

  If this doesn't help then please post your Datatables JS code and an example of your JSON response from the browser's network tools. Or use the debugger to collect this information. Post the debugger link here.

  Kevin
• SR27 Posts: 3Questions: 1Answers: 0

  July 2018

  Thanks. Here's a little more information. I think I am using the columns.data correctly, and I have made sure that my table columns match the number of columns in the js. I'm a bit stumped. I think the issue is with the formatting of the JSON, but I'm not sure how to output it differently.

  My html:

  <table id='tblPersonResults' class='table table-striped'>
      <thead>
        <tr>
           <th>Id</th>
            <th>First</th>
            <th>Last</th>
            <th>Other</th>
            <th>City</th>
            <th>Relationship</th>
         </tr>
        </thead>
        <tbody></tbody>
  </table>
  

  My js:

   var tblPersonResults = $('#tblPersonResults').DataTable({
              'ajax': {
                  "type": "POST",
                  "url": "my_url",
                  "dataType": 'JSON',
                  "contentType": "application/json; charset=utf-8",
                  "data": {},
                  "dataSrc": "d"
              },
              "columns": [
                  { "data": "id" },
                  { "data": "fName" },
                  { "data": "lName" },
                  { "data": "other" },
                  { "data": "address" },
                  { "data": "relationship" }
              ],
          });
  

  The JSON (output from vb.net with JsonConvert.SerializeObject method. Microsoft .net adds the "d"):

  {
     "d":[
        {
           "id":31,
           "fName":"tester",
           "lName":"testerlady",
           "other":"SSN: 1234",
           "address":"New Orleans, LA",
           "relationship":24
        },
        {
           "id":32,
           "fName":"tester",
           "lName":"testerman",
           "other":"SSN: 9876",
           "address":"New York, NY",
           "relationship":25
        }
     ]
  }
  
  
• SR27 Posts: 3Questions: 1Answers: 0

  July 2018

  Ok, I fixed it. The JSON was not getting processed correctly, so I had to wrap the dataSrc in a function like so:

  "dataSrc": function (json) { return $.parseJSON(json.d); }