Question

谁知道jqgrid subGridRowExpanded出来的子表格如何获取选中的行？

Answer 1

subGridRowExpanded: function(subgrid_id, row_id) {
// we pass two parameters
// subgrid_id is a id of the div tag created whitin a table data
// the id of this elemenet is a combination of the "sg_" + id of the row
// the row_id is the id of the row
// If we wan to pass additinal parameters to the url we can use
// a method getRowData(row_id) - which returns associative array in type name-value
// here we can easy construct the flowing
var subgrid_table_id, pager_id;
subgrid_table_id = subgrid_id+"_t";
pager_id = "p_"+subgrid_table_id;
$("#"+subgrid_id).html("<table id='"+subgrid_table_id+"' class='scroll'</table<div id='"+pager_id+"' class='scroll'</div");
jQuery("#"+subgrid_table_id).jqGrid({
url:"subgrid.php?q=2&id="+row_id,
datatype: "xml",
colNames: ['No','Item','Qty','Unit','Line Total'],
colModel: [
{name:"num",index:"num",width:80,key:true},
{name:"item",index:"item",width:130},
{name:"qty",index:"qty",width:70,align:"right"},
{name:"unit",index:"unit",width:70,align:"right"},
{name:"total",index:"total",width:70,align:"right",sortable:false}],
rowNum:20,

Answer 2

function getSelectedRows1() {
var grid = $("#jqGridPager");
return grid.getGridParam("selarrrow");
}