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Home/ Questions/Q 8784285
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Editorial Team
  • 0
Editorial Team
Asked: June 13, 20262026-06-13T21:02:55+00:00 2026-06-13T21:02:55+00:00

I have created a form based on JSON feed, when the user click on

  • 0

I have created a form based on JSON feed, when the user click on “Save”, I have bootstrap modal window showing up and form doesnt get submitted

But I noticed when I add this piece of code, the form is getting submitted, I need this to populated the modal window and display a table in the modal window.

How can i prevent it from submitting?
Modified code:

 $('#myModal').modal('show'); 
     $.each(data, function(key, val) {
        var tablefeed= $('<tr><td>ID: '+key+'</td><td id="'+key+'">'+val+'</td><tr>').appendTo('#display');
    });
$(".modal-body").html(tablefeed); 

I moved to ev.preventdefault as the @pmandell had suggested and that worked

      <script type="text/javascript">





   $('#myform').on('submit', function(ev) {

        ev.preventDefault();

    var data = $(this).serializeObject();
    json_data = JSON.stringify(data);
     $("#results").text(json_data); 
     $('#myModal').modal('show'); 
         $.each(data, function(key, val) {
            var tablefeed= $('<tr><td>ID: '+key+'</td><td id="'+key+'">'+val+'</td><tr>').appendTo('#display');
        });
    $(".modal-body").html(tablefeed); 





    }); 



        });
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-13T21:02:57+00:00Added an answer on June 13, 2026 at 9:02 pm
    $('#myform').on('submit', function(ev) {
        ev.preventDefault();
    ...
    
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