ajax.call 返回值怎么让它显示?

2016-07-07 15:04 来源:www.chinab4c.com 作者:ecshop专家

function merge_to_order()
{
var toOrderSn = document.getElementById('to_order_sn').value.substring(15);
Ajax.call('order_test.php?order_id=', toOrderSn, mergeto_orderResponse, 'POST', 'JSON');
}
怎么让这个mergeto_orderResponse显示在页面上?

回答:
function mergeto_orderResponse(result){
alert(result.参数);
}
result是返回的对象