获取jqgrid有多条行数据

选中一行可以用这样获得:

var selectedId = $("#manager_list")jqGrid("getGridParam", "selrow");

var rowData = $("#manager_list")jqGrid("getRowData", selectedId);

var selectedIds = $("#manager_list")jqGrid("getGridParam", "selarrrow");

先通过这一句获得所有选中的ids, 然后根据ids得到所有选中的记录

colNames=$("#UsersGrid")jqGrid('getGridParam','colNames')colModel=$("#UsersGrid")jqGrid('getGridParam','colModel') 楼主去学下调试工具，firebug或者chrome的开发工具，看结构就知道了

子表格也是一个jqgrid，你可以像 *** 作主表格一样 *** 作子表格的

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:"subgridphpq=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,

pager: pager_id,

sortname: 'num',

sortorder: "asc",

然后子表格的 *** 作就是jQuery("#"+subgrid_table_id)jqGrid('getGridParam','selarrrow');

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