TypeError：undefined不是jqGrid中的函数

我正在研究jqGrid。 我是新手。 我使用的数据类型为“本地”。 每当我以前执行行jQuery（“＃list2”）。jqGrid（{…}）时，我得到一个错误，因为’TypeError：undefined不是函数’没有其他信息我得到这个。 代码如下：

          $(document).ready(function () { var mydata = [ { id: "1", invdate: "2007-10-01", name: "test", note: "note", amount: "200.00", tax: "10.00", total: "210.00" }, { id: "2", invdate: "2007-10-02", name: "test2", note: "note2", amount: "300.00", tax: "20.00", total: "320.00" }, { id: "3", invdate: "2007-09-01", name: "test3", note: "note3", amount: "400.00", tax: "30.00", total: "430.00" }, { id: "4", invdate: "2007-10-04", name: "test", note: "note", amount: "200.00", tax: "10.00", total: "210.00" }, { id: "5", invdate: "2007-10-05", name: "test2", note: "note2", amount: "300.00", tax: "20.00", total: "320.00" }, { id: "6", invdate: "2007-09-06", name: "test3", note: "note3", amount: "400.00", tax: "30.00", total: "430.00" }, { id: "7", invdate: "2007-10-04", name: "test", note: "note", amount: "200.00", tax: "10.00", total: "210.00" }, { id: "8", invdate: "2007-10-03", name: "test2", note: "note2", amount: "300.00", tax: "20.00", total: "320.00" }, { id: "9", invdate: "2007-09-01", name: "test3", note: "note3", amount: "400.00", tax: "30.00", total: "430.00" } ]; jQuery("#list2").jqGrid({ datatype: "local", data: mydata, colNames: ['Inv No', 'Date', 'Client', 'Amount', 'Tax', 'Total', 'Notes'], colModel: [ { name: 'id', index: 'id', width: 60, sorttype: "int" }, { name: 'invdate', index: 'invdate', width: 90, sorttype: "date" }, { name: 'name', index: 'name', width: 100 }, { name: 'amount', index: 'amount', width: 80, align: "right", sorttype: "float" }, { name: 'tax', index: 'tax', width: 80, align: "right", sorttype: "float" }, { name: 'total', index: 'total', width: 80, align: "right", sorttype: "float" }, { name: 'note', index: 'note', width: 150, sortable: false } ], rowNum: 10, rowList: [10, 20, 30], jsonReader: { repeatitems: false }, //pager: '#pager2', sortname: 'id', viewrecords: true, sortorder: "desc", caption: "JSON Example" }); for (var i = 0; i <= _data.length; i++) jQuery("#list2").jqGrid('addRowData', i + 1, mydata[i]); }); 

上面的代码取自jqGrid官方网站

哪里出错了？ 另外，我应该使用local或json的数据类型，为什么？ 我在任何地方都没有找到更好，更容易理解的描述。