问问题
150 次
1 回答
0
因为 AJAX 是异步的。您需要放入$("#list").jqGrid({...
成功回调中:
// No need to define the variables outside
$.post('procstring.php', { data: window.location.hash },
function(result)
var grade = result.grade;
var section = result.section;
var from = result.from;
var until = result.until;
var user = result.user;
$("#list").jqGrid({...
},
'json');
于 2010-08-26T09:28:25.370 回答