Answers

• colin Posts: 15,176Questions: 1Answers: 2,589

  July 2018

  Hi @Barbaros ,

  It depends on whether you're using arrays or objects. You're using an array with you add the row (row.add()), so you need to use the index for the position of each field. If you were using objects, as in the first example on that page, then you could use the element name.

  Cheers,

  Colin
• Barbaros Posts: 3Questions: 1Answers: 0

  July 2018

  Hi @colin ,thanks for your reply.
  I got it .
  one more question- (without ajax): this datatable is inside form and when i submit the form i want to post all table data with hidden input. is it possible or is there any better way to pass all data ?
  Thanks .

  javascript:
  $(this).submit( function() {
   
          var table = $('#example').DataTable();
          var data = t.rows().data().toArray();
          $('#dt_value_1').val(data);
  
      } );
  

  and html :

  <input type="hidden" name="dt_value[]" id="dt_value_1" class="form-control" autocomplete="off">
  

  and php:

  $_POST['dt_value']
  

  it works but it merge all data in one row .
  "[\"messi,123456789,messi@messi.com,ronaldo,111444777,ronaldo@ronaldo.com\"]"
• colin Posts: 15,176Questions: 1Answers: 2,589

  July 2018 Answer ✓

  Hi @Barbaros ,

  You could iterate over the rows one at a time, and add it to an array, something like this pseudo code:

  var data = [];
  table.rows().every(function() {
    data.push(this.data().toArray())
  }
  

  That would make a 2d array which would be tidier.

  Cheers,

  Colin
• bindrid Posts: 730Questions: 0Answers: 119

  July 2018 Answer ✓

  Try JSON.stringify to serialized the data before you put it in the hidden. Then deserialize on the back end. That should give you your array to work with.
• Barbaros Posts: 3Questions: 1Answers: 0

  July 2018

  @bindrid , @colin thanks for your replies,
  after i changed

  name="dt_value[] to name="dt_value
  

  and add using JSON.stringify to serialized the data

   var arry=t.rows().data().toArray();
          var data = JSON.stringify(arry);
          $('#dt_value_1').val(data);
  

  and for backend

   $response = $this->input->post('dt_value');
   $myjson = json_decode($response);
  
  

  is worked for me, thanks for your helps .