错误& # 39;美元.类型不是函数var obj = $.类型(pref);'在jqGrid中提交表单时

最近，我正在使用jqGrid来表示我的业务数据集，并且我发现jqGrid具有编辑/添加/删除/查看功能，因此我尝试使用它。但是我遇到了一个问题，尽管我尽力了，我还是不能解决它。

我这样做的例子从这个网站:

jQuery("#deptGrid").jqGrid('navGrid', '#deptNav', {
    view : true
}, 
{
    jqModal : true,
    checkOnUpdate : true,
    savekey : [ true, 13 ],
    navkeys : [ true, 38, 40 ],
    checkOnSubmit : true,
    reloadAfterSubmit : false,
    closeOnEscape : true,
    bottominfo : "Fields marked with (*) are required"
}, {
    jqModal : true,
    checkOnUpdate : true,
    savekey : [ true, 13 ],
    navkeys : [ true, 38, 40 ],
    checkOnSubmit : true,
    reloadAfterSubmit : false,
    closeOnEscape : true,
    bottominfo : "Fields marked with (*) are required"
}, {
    reloadAfterSubmit : false,
    jqModal : false,
    closeOnEscape : true
}, {
    closeOnEscape : true
}, {
    navkeys : [ true, 38, 40 ],
    height : 250,
    jqModal : false,
    closeOnEscape : true
});

UPDATE:以下是我的全部内容:

<link rel="stylesheet" type="text/css" href="<%=BASE_PATH %>admin/styles/stylesheet.css" />
<link rel="stylesheet" type="text/css" href="<%=BASE_PATH %>admin/js/jquery/ui/themes/ui-lightness/ui.all.css" />
<link rel="stylesheet" type="text/css" media="screen" href="<%=BASE_PATH %>jqGrid4.4.1/themes/redmond/jquery-ui-1.8.2.custom.css" />
<link rel="stylesheet" type="text/css" media="screen" href="<%=BASE_PATH %>jqGrid4.4.1/themes/ui.jqgrid.css" />
<link rel="stylesheet" type="text/css" media="screen" href="<%=BASE_PATH %>jqGrid4.4.1/themes/ui.multiselect.css" />
<!-- Including JS -->
<script type="text/javascript" src="<%=BASE_PATH %>admin/js/jquery/jquery.min.js"></script>
<script type="text/javascript" src="<%=BASE_PATH %>admin/js/jquery/jquery-ui-1.8.custom.min.js"></script>
<script type="text/javascript" src="<%=BASE_PATH %>admin/js/jquery/superfish/js/superfish.js"></script>
<script type="text/javascript" src="<%=BASE_PATH %>admin/js/jquery/tab.js"></script>
<script type="text/javascript" src="<%=BASE_PATH %>admin/js/global.js"></script>
<script type="text/javascript" src="<%=BASE_PATH %>admin/js/jquery/thickbox/thickbox.js"></script>
<link  rel="stylesheet" type="text/css" href="<%=BASE_PATH %>admin/js/jquery/thickbox/thickbox.css" />
<script type="text/javascript" src="<%=BASE_PATH %>admin/js/jquery/jquery.form.js"></script>
<script type="text/javascript" src="<%=BASE_PATH %>admin/js/jquery/jquery.validate.js"></script>
<script type="text/javascript" src="<%=BASE_PATH %>admin/js/jquery/jquery-ui-timepicker-addon.js"></script>
<script src="<%=BASE_PATH %>admin/js/jquery/ui/jquery.bgiframe-2.1.2.js" type="text/javascript"></script>
<script src="<%=BASE_PATH %>admin/js/jquery/ui/jquery.ui.core.js" type="text/javascript"></script>
<script src="<%=BASE_PATH %>admin/js/jquery/ui/jquery.ui.widget.js" type="text/javascript"></script>
<script src="<%=BASE_PATH %>admin/js/jquery/ui/jquery.ui.mouse.js" type="text/javascript"></script>
<script src="<%=BASE_PATH %>admin/js/jquery/ui/jquery.ui.button.js" type="text/javascript"></script>
<script src="<%=BASE_PATH %>admin/js/jquery/ui/jquery.ui.draggable.js" type="text/javascript"></script>
<script src="<%=BASE_PATH %>admin/js/jquery/ui/jquery.ui.position.js" type="text/javascript"></script>
<script src="<%=BASE_PATH %>admin/js/jquery/ui/jquery.ui.resizable.js" type="text/javascript"></script>
<script src="<%=BASE_PATH %>admin/js/jquery/ui/jquery.ui.dialog.js" type="text/javascript"></script>
<script src="<%=BASE_PATH %>admin/js/jquery/ui/jquery.effects.core.js" type="text/javascript"></script>
<script src="<%=BASE_PATH %>js/kevinshi.js" type="text/javascript"></script>
<script src="<%=BASE_PATH %>jqGrid4.4.1/js/jquery.jqGrid.src.js" type="text/javascript"></script>
<script src="<%=BASE_PATH %>jqGrid4.4.1/js/i18n/grid.locale-en.js" type="text/javascript"></script>

更新:每当我点击编辑/添加/删除按钮时，它总是在firebug中引起错误:

$.type is not a function 
var obj = $.type( pref );

我将其跟踪到jqGrid源代码中，这个错误发生在文件jquery.jqGrid.src.js的第52行，如下所示:

stripPref : function (pref, id) {
var obj = $.type( pref ); // Happened at this line.
if( obj == "string" || obj =="number") {
pref = String(pref);
id = pref !== "" ? String(id).replace(String(pref), "") : id;
}
return id;
}, 

所以，就在这一刻，我可以说找到了jQuery.type()方法，但是我已经包含了库jquery.min.js，我很困惑。