Empty (val) html field with ajax and enter next value in this field from database

I am working on a student management system, which is why I have so many fields in my HTML form, including the following

  • Student ID
  • Paper number
  • Assessment

HTML FORM

Ajax is used to send data to the form, so I would like to see the form is submitted to empty (val) student id field and fill the student id field with next student id from MySQL

, as well as field Val Score.

My problem is how to fill the student id field with the following value from MySQL.

If this is not submitted via Ajax, I would just fetch and repeat the next value from the MySQL database.

Below is the Ajax am am code to submit the form

<script type="text/javascript">
    $(document).ready(function() {
        $("#keyingForm").submit(function() {
            $("#loading").slideDown();
                 var frm = $('#keyingForm');
                 $.ajax({
                     type: frm.attr('method'),
                     url: 'scoring.php',
                     data: frm.serialize(),
                     success: function (data) {
                         $('#loadData').html(data);
                         $("#loading").slideUp();
               }, error: function(jqXHR, textStatus, errorThrown){
              console.log(" The following error occured: "+ textStatus, errorThrown );
            } });
           return false;
        });
    });     
</script>

      

Student id studentid

and grade for field scoreid

I hope this makes sense as I am not very good at Ajax or JavaScript, thanks for your hand.

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1 answer


On your PHP project, try loading the following js



<script type="text/javascript">
    //$(document).ready(function() {
        $('#studentid').val(''); $('#x1').val(''); 
        var tav    = $('#studentid').val(),
        strPos     = $('#studentid')[0].selectionStart;
        front      = (tav).substring(0,strPos),
        back       = (tav).substring(strPos,tav.length); 
        $('#studentid').val(front + "<?php echo $newstudentid; ?>" + $(this).text() + '' + back);               
    //});
</script>

      

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