Javascript jqxGrid失败,出现错误;对象[Object Object]没有方法';jqGrid'&引用;

Javascript jqxGrid失败,出现错误;对象[Object Object]没有方法';jqGrid'&引用;,javascript,jquery,jqxgrid,Javascript,Jquery,Jqxgrid,我正在尝试将jqxgrid嵌入到我的HTML页面中 以下是我导入的库: <script src="./wicket/resource/org.apache.wicket.resource.JQueryResourceReference/jquery/jquery-1.10.1-ver-1379671500000.js"></script> <script src="./wicket/resource/org.apache.wicket.ajax.AbstractDe

我正在尝试将jqxgrid嵌入到我的HTML页面中

以下是我导入的库:

<script src="./wicket/resource/org.apache.wicket.resource.JQueryResourceReference/jquery/jquery-1.10.1-ver-1379671500000.js"></script>
<script src="./wicket/resource/org.apache.wicket.ajax.AbstractDefaultAjaxBehavior/res/js/wicket-event-jquery-ver-1379671500000.js"></script>
<script src="./wicket/resource/org.apache.wicket.ajax.AbstractDefaultAjaxBehavior/res/js/wicket-ajax-jquery-ver-1379671500000.js"></script>
<script src="/jquery/jquery-ui.min.js"></script>
<script src="/jqwidgets/jqxcore.js"></script>
<script src="/jqwidgets/jqxdata.js"></script>
<script src="/jqwidgets/jqxbuttons.js"></script>
<script src="/jqwidgets/jqxscrollbar.js"></script>
<script src="/jqwidgets/jqxmenu.js"></script>
<script src="/jqwidgets/jqxcheckbox.js"></script>
<script src="/jqwidgets/jqxlistbox.js"></script>
<script src="/jqwidgets/jqxdropdownlist.js"></script>
<script src="/jqwidgets/jqxgrid.js"></script>
<script src="/jqwidgets/jqxgrid.columnsresize.js"></script>
<script src="/jqwidgets/jqxgrid.edit.js"></script>
<script src="/jqwidgets/jqxgrid.filter.js"></script>
<script src="/jqwidgets/jqxgrid.pager.js"></script>
<script src="/jqwidgets/jqxgrid.selection.js"></script>
<script src="/jqwidgets/jqxgrid.sort.js"></script>
我再次省略了(在我看来)不必要的细节,因为jqxgrid的初始化工作得很好,我看到了包含我希望它显示的所有数据的表

但当我编辑单元格并结束编辑时,行

$('#jqxgrid').jqGrid('getCell',args.rowindex,'Name')
在控制台中产生错误:

TypeError: Object [object Object] has no method 'jqGrid'
我做了一些研究,但无法做出明确的解释。有什么不对劲吗?有什么我忘了的吗?顺序有问题吗?

键入错误,更改

$('#jqxgrid').jqGrid(... 


它不应该是
$('#jqxgrid').jqxgrid('getCell',args.rowindex,'Name')?尴尬。。。未经校对不得复制粘贴。谢谢
$('#jqxgrid').jqGrid(... 
$('#jqxgrid').jqxGrid(...