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form 表單 ajaxSubmit 提交后接收不到返回值

Java

Cats萌萌 2018-08-09 08:09:39

前臺代碼$("#formSave").ajaxSubmit({success: function (data) {ShowMsg(data.message)if (data.success) {window.location.href = "/CouponManage/BusinessCoupon/List";}},error: function (data) {ShowMsg(data.message);}});后臺代碼public JsonResult UpdateSave(){Guid guid = new Guid(Request["CouponGuid"]);string Title = Request["title"];string Detial = Request["desc"];string msg = "";CouponLogic bll = new CouponLogic(this.OpertorGuid);Coupon coup = new Coupon();coup.Guid = guid;coup.Title = Title;coup.Description = Detial;bool flag = bll.UpdateCoupon(UserAccount, ChainStoreGuid, coup, out msg);return Json(new { success = flag, msg = msg });}結(jié)果直接在瀏覽器里面輸出了 {"success":true,"message":"修改成功!"} ajax提交哪里沒有辦法處理結(jié)果

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1 回答

RISEBY

TA貢獻(xiàn)1856條經(jīng)驗獲得超5個贊

  $("#formSave").ajax({
            type: 'post', // 提交方式 get/post
            url: 'your url', // 需要提交的 url            data: {                'title': title,                'desc': desc,                'CouponGuid':CouponGuid
            },
            success: function(data) { // data 保存提交后返回的數(shù)據(jù)，一般為 json 數(shù)據(jù)                // 此處可對 data 作相關(guān)處理
                alert('提交成功！');
            }
            $(this).resetForm(); // 提交后重置表單
        });

反對回復(fù) 2018-08-27