jquery的subgrid如何自动展开

[momochaasdfasdf]

请教各位大神啊，急啊
jquery的subgrid如何自动展开？

代码如下：
$(function(){
// 配置jqGrid组件
$("#gridTable").jqGrid({
url: "jqGrid05.action",
datatype: "json",
mtype: "GET",
height: 350,
width: 600,
colModel: [
{name:"id",index:"id",label:"编码",width:40},
{name:"lastName",index:"lastName",label:"姓",width:80},
{name:"firstName",index:"firstName",label:"名",width:80},

{name:"email",index:"email",label:"电子邮箱",width:160,sortable:false},
{name:"telNo",index:"telNo",label:"电话",width:120,sortable:false}
],
viewrecords: true,
rowNum: 15,
rowList: [15,50,100],
prmNames: {search: "search"},
jsonReader: {
root:"gridModel",
records: "record",
repeatitems : false,
},
pager: "#gridPager",
caption: "联系人列表",
hidegrid: false,
shrikToFit: true,
subGrid: true, // (1)开启子表格支持
subGridRowExpanded: function(subgrid_id, row_id) { //
(2)子表格容器的id和需要展开子表格的行id，将传入此事件函数
var subgrid_table_id;
subgrid_table_id = subgrid_id + "_t"; //
(3)根据subgrid_id定义对应的子表格的table的id

var subgrid_pager_id;
subgrid_pager_id = subgrid_id + "_pgr" //
(4)根据subgrid_id定义对应的子表格的pager的id

// (5)动态添加子报表的table和pager
$("#" + subgrid_id).html("<table id='"+subgrid_table_id+"'
class='scroll'></table><div id='"+subgrid_pager_id+"'
class='scroll'></div>");

// (6)创建jqGrid对象
$("#" + subgrid_table_id).jqGrid({
url: "fetchPatentCases.action?contact.id="+row_id, //
(7)子表格数据对应的url，注意传入的contact.id参数
datatype: "json",
colNames: ['编号','内部编码','名称','申请号'],
colModel: [
{name:"id",index:"id",width:80,key:true},
{name:"internalNo",index:"internalNo",width:130},
{name:"name",index:"name",width:80,align:"right"},

{name:"applicationNo",index:"applicationNo",width:80,align:"right"}
],
jsonReader: { // (8)针对子表格的jsonReader设置
root:"gridModel",
records: "record",
repeatitems : false
},
prmNames: {search: "search"},
pager: subgrid_pager_id,
viewrecords: true,
height: "100%",
rowNum: 5
});
}
});

});