Replies

• Tom (DataTables) Posts: 139Questions: 0Answers: 26

  May 2016

  Hi Bob

  I believe this should fix it.

  I changed
  "data": testdata,
  to
  "data": testdata.data,

  Full snippet below

    var table = $('#example2').DataTable( {
                  "data": testdata.data,
                  "columns": [
                      {
                          "className":      'details-control',
                          "orderable":      false,
                          "data":           null,
                          "defaultContent": ''
                      },
                      { "data": "ID" },
                      { "data": "CAGE" },
                      { "data": "Name" }
                  ],
                  "order": [[1, 'asc']]
              } );
  
• Bob Rodes Posts: 17Questions: 1Answers: 0

  May 2016 edited May 2016

  And there's my data! Fascinating. Thanks a lot! After looking a little deeper, I can see why we have to refer to the wrapper data object here, although why we don't have to with the ajax/json file call (which of course we couldn't really: I can't see where ajax: 'json.txt'.data is anything remotely like correct syntax) is a bit obscure to me. Perhaps you have some insight on that you can share.

  I'm presently using PHP calls to get data in object variables, pass them directly into JS object variables, and pass those into DataTableI() calls. I'm working to get Allen's AJAX-based example to run this way. Now, the PHP data in the call is a simple array of objects, in [ {},{},{etc} ] format. This format appears to require that we wrap that in another object: {data:[ {},{},{etc} ] }. I couldn't get any of my test examples of my own data to work until I did this, anyway. Is this required because of the way the child data are set up? Here's Allen's code to show the children:

      // Add event listener for opening and closing details
      $('#example tbody').on('click', 'td.details-control', function () {
          var tr = $(this).closest('tr');
          var row = table.row( tr );
   
          if ( row.child.isShown() ) {
              // This row is already open - close it
              row.child.hide();
              tr.removeClass('shown');
          }
          else {
              // Open this row
              row.child( format(row.data()) ).show();
              tr.addClass('shown');
          }
      } );
  

  In particular, the line row.child( format(row.data()) ).show(); obviously requires the wrapping data object. But I'd like to avoid editing the incoming array in my own app to wrap it if I can. I tried the naive approach of just removing the wrapper, and just using row.child( format(row()) ).show(); which was, well, naive: of course it blew up in spectacular fashion.

  So, is there a way to avoid wrapping the object in the data:{} object, or do I have to have an object within an object to make this work? In any case, thanks a lot for getting me going again.
• allan Posts: 63,498Questions: 1Answers: 10,471 Site admin

  May 2016

  I'm going to step in for just a second and answer your first query, but leave your follow up question to Tom.

  can see why we have to refer to the wrapper data object here, although why we don't have to with the ajax/json file call (which of course we couldn't really: I can't see where ajax: 'json.txt'.data is anything remotely like correct syntax) is a bit obscure to me. Perhaps you have some insight on that you can share.

  The wrapper object when getting the data via Ajax is a legacy decision of mine. It was (and still use) useful to allow the JSON to carry more than just the data needed for the DataTable. However, it is quite possible to change the data array's location, or even to do away with the wrapper object - the ajax.dataSrc option provides that ability.

  Ultimately you just need to get the data array into the table - data does that directly, while ajax provides some mapping options.

  This may or may not answer your follow up question - let us know!

  Allan
• Bob Rodes Posts: 17Questions: 1Answers: 0

  May 2016

  Thanks for your explanation, Allan, that all makes sense. Once I understood what was going on, I found it simple to wrap a the data: object around the incoming data ($rs is the PHP variable containing the JSON data):

              var theData = { "data": <?php echo $rs; ?> } 
              var table = $('#example2').DataTable( {
                  "data": theData.data,
                  [etc.]
  

  This is all very helpful insight, as I have been resorting with other tables to transforming the incoming object data into straight array data. I'll have to go back now and redo that.