Jquery的post请求

 

 

var _param = {
"code": code,
"IsReturn": isReturn,
"type": "isReturn"
};

$.ajax({
type: "post",
url: "UrlAPI",
data: _param,
success: function (result) {
//console.log(result);
}});

posted @ 2022-03-15 13:46  艾特-天空之海  阅读(542)  评论(0编辑  收藏  举报