[HttpPost]
public JsonResult DelayReminder(string reminderId)
{
ReminderStatus rs = new ReminderStatus();
rs.BaseProps.RequesterUserInfo.UserID = SessionManager.Current.CurrentUser.UserID;
ReminderServiceHelper.InsertNewStatus(Convert.ToInt32(reminderId), rs);
return Json(apptype, JsonRequestBehavior.AllowGet); // Problem...
}
而不是返回Json(apptype,JsonRequestBehavior.AllowGet); 我如何在下面写?
return RedirectToAction("GetLawDetail", "Law", new { _lawID = baseappid });
如果有人想看Javascript:
$.ajax({
type: 'POST',
url: '/Reminders/DelayReminder/',
data: {
'apptype': '@ViewData["apptype"]',
'baseappid': '@ViewData["baseappid"]',
'reminderId': $("#reminderId").val(),
'ddlDelayDuration': $("#ddlDelayDuration").val()
},
dataType: 'json',
success: function (result) {
if (result != null) {
}
....
..
如何在jsonresult内部返回 Law controller to GetLawDetail actionresult ?
答案 0 :(得分:0)
您可以退回此操作:
return GetLawDetail(baseappid)
您还应该将返回类型更改为@ im1dermike指向的Action结果。
这里是完整的代码:
[HttpPost]
public ActionResult DelayReminder(string reminderId)
{
ReminderStatus rs = new ReminderStatus();
rs.BaseProps.RequesterUserInfo.UserID = SessionManager.Current.CurrentUser.UserID;
ReminderServiceHelper.InsertNewStatus(Convert.ToInt32(reminderId), rs);
return GetLawDetail(apptype); // Problem...
}
虽然,它会返回整个渲染页面。