Answers

• DAustin Posts: 16Questions: 0Answers: 1

  October 2018

  As you are using pagination, the form cannot "see" the pages that aren't displayed, the html is removed from the DOM.

  Option 1, update the pageLength of the table to -1 (show all) when submitting:

  var table = $('#example').DataTable();
  $("#form").on("submit", function(e){
      //Stop the form submitting
      e.preventDefault();
      //Show all the rows
      table.page.len( -1 ).draw();
      //Submit the form now it can see everything
      $(this).submit();
  });
  

  Option 2, Store the values in your own hidden input:

  //Array to hold the checked ids
  var checkboxes = []
  
  function updateCheckboxes(checkbox){
      //Get the row id
      var id = checkbox.data("id");
      //Check the array for the id
      var arrPos = checkboxes.indexOf(id);
      //If it exists and we unchecked it, remove it
      if(arrPos > -1 && !checkbox.checked){
          checkboxes.splice(arrPos,1);
      }
      //Else it doesn't exist and we checked it
      else if (checkbox.checked)
      {
          checkboxes.push(id);
      }
      //Finally update the hidden input
  $("#idOfHiddenInput").val(checkboxes);
  }
  
  //Event listener to detect changes
  $("body").on("change",".checkbox-class", updateCheckboxes($(this)))
  
  

  The advantage of option 1 is it's quick, but it will look ugly as the table suddenly grows in size. Option 2 requires you to handle the hidden values in the form action, but won't cause any visual issues.
• colin Posts: 15,142Questions: 1Answers: 2,586

  October 2018

  Hi @amithks ,

  I don't quite follow your process - i.e. how would you check a checkbox on a different page, and when would it be submitted. Are you using Editor?

  We're happy to take a look, but it would help, as per the forum rules, if you could link to a running test case showing the issue so we can offer some help. Information on how to create a test case (if you aren't able to link to the page you are working on) is available here.

  Cheers,

  Colin
• DAustin Posts: 16Questions: 0Answers: 1

  October 2018

  I think this is actually a bad design choice. When a user uses a form they're going to expect all the inputs to be visible on that page, or across several pages as they go through the steps. By using pagination to hide inputs there's a strong chance the user will not see them either through laziness or obliviousness.

  I say this because my colleagues often request that a table has all the results on one page that they can scroll through, as it can be annoying for them to compare values on different pages.

  If you show everything on the first page, no one can claim they missed the information/input by it being hidden, plus you'll save yourself some coding time
• amithks Posts: 2Questions: 1Answers: 0

  October 2018

  Thanks for the solution DAustin.