Javascript 如何在jquery排序表中获取移动的ID和替换的ID?

Javascript 如何在jquery排序表中获取移动的ID和替换的ID?,javascript,jquery,jquery-ui,jquery-ui-sortable,Javascript,Jquery,Jquery Ui,Jquery Ui Sortable,我正在使用jquerysortable。我希望能够获取移动项目的ID和替换项目的ID。到目前为止,我能够获取移动元素的ID,但无法获取它所替换的元素的ID 我的代码是: $(function () { $("#sortable").sortable({ stop: function (event, ui) { var moved = ui.item, replaced = ui.item.prev();

我正在使用jquerysortable。我希望能够获取移动项目的ID和替换项目的ID。到目前为止,我能够获取移动元素的ID,但无法获取它所替换的元素的ID

我的代码是:

$(function () {
    $("#sortable").sortable({
        stop: function (event, ui) {
            var moved = ui.item,
                replaced = ui.item.prev();

            // if replaced.length === 0 then the item has been pushed to the top of the list
            // in this case we need the .next() sibling
            if (replaced.length == 0) {
                replaced = ui.item.next();
            }

            alert("moved ID:" + moved.attr("id"), "replaced ID:" + replaced.attr("id"));
        }
    });
});
如何获取被替换元素和被移动元素的ID


实际上它是有效的,您只是用错误的参数调用alert;将其替换为
console.log
,或按如下方式连接字符串:

alert("moved ID:" + moved.attr("id") + "replaced ID:" + replaced.attr("id"));

(我用
+
替换了

实际上它是有效的,你只是用错误的参数调用alert;将其替换为
console.log
,或按如下方式连接字符串:

alert("moved ID:" + moved.attr("id") + "replaced ID:" + replaced.attr("id"));
(我将
替换为
+