Answers

• tangerine Posts: 3,365Questions: 39Answers: 395

  September 2017

  https://jsonlint.com/ says your JSON is invalid....

  manageAtmFleetTable = $("#manageAtmFleetTable").DataTable({
  'ajax' : 'php_action/fetchTerminal.php',
  'order' : [],
  });
  

  ....and that's not HTML.
• kamar Posts: 7Questions: 1Answers: 0

  September 2017 edited September 2017

  Thanks for your response tangerine, however the JSON i posted is the data i want passed from my DB to the table on my website.

  i know the below is javascript, however, that is how i am passing the data to my website table

  manageAtmFleetTable = $("#manageAtmFleetTable").DataTable({
  'ajax' : 'php_action/fetchTerminal.php',
  'order' : [],
  });
  

  here is fetchTerminal.php

  <?php
  
  require_once 'core.php';
  
  $sql = "SELECT atm_id, terminal_id, terminal_name, branch_code, 
  onsite_offsite, vendor_FK, local_ip, custodian_name, default_gateway, 
  region FROM atm_database ";
  
  $result = $connect->query($sql);
  
  $output = array('data' => array());
  
  if($result->num_rows > 0) {
      while($row = $result->fetch_array()) {
           // echo "<pre>";
           // print_r($row);
           // echo "</pre>";
          $atmId = $row[0];
          
  
  
  
      $output['data'][] = array( 
              $atmId,
              $row[1],
              $row[2],
              $row[3],
              $row[4],
              $row[5],
              $row[6],
              $row[7],
              $row[8],
              $row[9]
  
              );
  
  
          
      } // while
      
  } // /if
  
  $connect->close();
  
  echo json_encode($output);
  

  Edited by Allan - Syntax highlighting. Details on how to highlight code using markdown can be found in this guide.
• kthorngren Posts: 21,449Questions: 26Answers: 4,975

  September 2017

  As tangerine mentioned your JSON data is not a valid JSON string. This page discusses the data structures Datatables supports.
  https://datatables.net/manual/data/

  The structure you posted does not meet the requirements.

  Kevin
• kamar Posts: 7Questions: 1Answers: 0

  September 2017

  Thank you kthorngren, i have gone through the document how ever i looked through my code and realize i was using ajax form.serialize(), pls could that be the problem

  if(terminalId && terminalName) {
  var form = $(this);

  //button loading
  $("#createTerminalBtn").button('loading');
  
  $.ajax({
      url: form.attr('action'),
      type: form.attr('method'),
      data: form.serialize(),
      dataType: 'json',
      success:function(response){
          //button loading
          $("#createTerminalBtn").button('reset');
  
          if(response.success == true) {
  
• allan Posts: 63,689Questions: 1Answers: 10,500 Site admin

  September 2017

  What is shown above as "JSON" appears to be the result of the print_r on line 16 (commented out at the moment) rather than the result of the echo json_encode(), which I think is causing some of the confusion.

  Could you run the debugger on your table so we can take a look at the raw response from the server please? There isn't anything obviously wrong that I can see in the code above.

  It would also be worth checking the server's error log to see if there is any information shown there.

  Thanks,
  Allan
• kamar Posts: 7Questions: 1Answers: 0

  September 2017

  Allan, thank you for your reply however i need clarification on where to paste the debugger code, google chrome has no debugger and mozilla debugger is not writeable. Please help
• tangerine Posts: 3,365Questions: 39Answers: 395

  September 2017

  You are expected to use DataTables' own debugger. Instructions for use are right there in the link Allan gave you.
• kamar Posts: 7Questions: 1Answers: 0

  September 2017

  Please what does this line on datatable.min.css 48 mean

  c);for(b=0;b<f.length;b++)M(a,f[b]);a.iInitDisplayStart=
• allan Posts: 63,689Questions: 1Answers: 10,500 Site admin

  September 2017

  Its a for loop that is part of the minimised code. I could explain it character by character, but I doubt that is actually want to want to know.

  As I said before, I'm happy to help, but please use the debugger I linked to and @tangerine explained, or link to a page showing the issue.

  Allan