Jquery的post请求

var _param = { "code": code, "IsReturn": isReturn, "type": "isReturn" }; $.ajax({ type: "post", url: "UrlAPI", data: _param, success: function (result
posted @ 2022-03-15 13:46  艾特-天空之海  阅读(541)  评论(0)  编辑  收藏  举报