PHP Ajax Call Navigating to Action Page on Form Submit

Question

I have this ajax call in my php code that still navigates to action page. I want the ajax call to stay on the same page and only update part of the page. What I'm doing wrong here:

<form method="post" id="holderSave" action="student_add_scene.php">
            <h3> Save to Holder</h3><br/>
            <div id="cutfrom">
                <input placeholder="Cut from" name="from" id="from">
                <button href="#" onclick="captureElapsed('elapseFrom', 'from');
                        return false;">capture</button>
            </div>

            <div id="cutto">  
                <input placeholder="Cut to" name="to" id="to" >
                <button href="#" onclick="captureElapsed('elapseTo', 'to');
                        return false">capture</button>
            </div>

            <div id="savecapt">
                <input placeholder="add label" id="label" name="label"/>
                <input type='submit' value='Save' class='button'>
            </div>
        </form>




<script>
        $('#holderSave').on("submit", (function (e) { // catch the form's submit event
            e.preventDefault();
            $.ajax({// create an AJAX call...
            var $form = $(this), url = $form.attr('action');
            var posting = $.post( url, { from: $('#from').val(), label:$('#label').val(), to: $('#to').val()});
            posting.done(function(response)){
                $('#holderSave').html(response); // update the DIV
                alert(response);
            }

        });
        return false;
        }));
    </script>

Answer 1

1. Change submit into button with id

   <input type='button' id="save" value='Save' class='button'>

2. Trigger click event for button, serialize the form data and post the data via ajax

   $(document).ready(function() { 
   
       $('#save').click(function(){ // catch the form's submit event
            var form = $('#holderSave');
            $.ajax( {
                 type: "POST",
                 url: form.attr( 'action' ),
                 data: form .serialize(),
                 success: function( response ) {
                   console.log( response );
                 }
           } );
       });
   });
   

Answer 2

This is a syntax error:

$.ajax({// create an AJAX call...
var $form = $(this), url = $form.attr('action');

You seem to be trying to treat the object literal you pass to ajax as a function body. It isn't, so you can't just write statements in it.

Since you make the ajax request with $.post later, $.ajax is entirely pointless. Remove that line and it should work.

---

Fixed code. Aside from the pointless half-a-call to .ajax, you had a bunch of syntax errors which I've fixed while reformatting it.

Use your browser's developer tools console. Use http://jshint.com/

// Remove pointless ( from before the second argument of the call to on().
$('#holderSave').on("submit", function(e) {
        e.preventDefault();
        // Remove half a call to .ajax
        var $form = $(this),
            url = $form.attr('action');
        var posting = $.post(url, {
            from: $('#from').val(),
            label: $('#label').val(),
            to: $('#to').val()
        });
        posting.done(function(response) {
            $('#holderSave').html(response);
            alert(response);
        // Remove line with extra } here
        });
        return false;
    // Remove extra ) here
    });