我可以使用summernote html代码传递给jQuery ajax调用

我正在尝试访问summernote的值，并尝试通过ajax调用传递它，但它没有调用该控制器的操作。

log.cshtml

<div id="dvUpdate" class="tab-pane fade">
                        <form class="log-activity-form" id="addUpdate" name="addUpdate" novalidate>
                            <div class="wrapper wrapper-content animated fadeInRight">
                                <div class="row">
                                    <div class="col-sm-2">
                                        <label class="control-label">Update</label>
                                    </div>
                                    <div class="col-sm-10">
                                        <textarea rows="8" cols="150" class="summernote input-block-level" id="update" required></textarea>
                                    </div>
                                </div>
                                <div class="row">
                                    <div class="col-sm-2">
                                        <label class="control-label">&nbsp;</label>
                                    </div>
                                    <div class="col-sm-10">
                                        <button class="btn btn-sm btn-primary pull-right" style="margin-top:10px;" type="submit"><strong>Save Update</strong></button>
                                    </div>
                                </div>
                            </div>
                        </form>
                    </div>

javascript

 $('#update').summernote({
        focus: true,
        height: 500,
    });

    $("#addUpdate").submit(function () {            
        var textareaValue = $('#summernote').summernote('code');           

        var activityData = { distance: "", distancetype: "", hours: "", minutes: "", seconds: "", activity: "Update", activityName: "", date: moment.utc(), notes: textareaValue };

        $.ajax({
            type: "POST",
            contentType: 'application/json',
            url: "/Runner/CreateActivityLog",
            data: JSON.stringify(activityData),
            success: function (data) {
                //refresh feeds section                   
            }               
        });
        return false;
    });

我不想使用剃刀语法来提交表单。我尝试了多种组合以获取summernote的值，即为object或null。

我尝试获取summernote代码的各种方法 var textareaValue = $('#summernote').code(); var content = $('textarea[name="Update"]').html($('#summernote').code()); var data = $('#summernote').summernote('code');