在Java中使用Ajax更改状态

我需要更改数据库中事件的状态。为此，我创建了带有下拉菜单的按钮，其中包含可以更改的所有状态。当我按下此按钮并选择某个变体时，它正在更改，但在我的事件字段“所有者、城市、地址”中变为空。怎么了

 function html_element(aData, status) {
        var parent_element = $("<a/>",
        {
            "href" : "#"
        }).click(aData, function(e) {
        aData.status = status;
        e.preventDefault();
        $.ajax({
        url : 'event-update',
        dataType : 'json',
        beforeSend : function() {
        },
        contentType : "application/json",
        "type" : "POST",
        data : JSON.stringify(aData),
        success : function(response) {
            $('td:eq(6)',nRow).html(response.status);
        }
        });
        });
        return parent_element;
        };
    var button = $(
            "<button/>",
            {
            text : "Action",
            "type" : "button",
            "data-toggle" : "dropdown",
            "class" : "btn btn-default dropdown-toggle",
        });
    function form_element(aData) {
            var approve = html_element(aData, "APPROVED");
            var pending = html_element(aData, "PENDING");
            var refuse = html_element(aData, "REFUSED");
            approve.html("Approve");
            pending.html("Pending");
            refuse.html("Refuse");
            var final_element = div.append(button);
           final_element.append(ul.append(li
            .append(approve)
           .append(pending)
            .append(refuse)));
            return final_element;
            }
            row = $('td:eq(7)', nRow).html(form_element(aData));
function change_request_status(id, element) {
            $.ajax({
                url : 'response-event',
                beforeSend : function() {
                console.log('start');
                element.style.display = "none";
                loader.insertAfter(element);
                },
            data : {
                "status" : status,
                "address": address,
                "owner" : owner,
                cityId : city
            },
            success : function(response) {
                console.log(data);
                element.nextSibling.style.display = "none";
                if (response == "Request send") {
                successRequest.insertAfter(element);
                } else {
                failRequest.insertAfter(element);
                }
            },
            });
            }