[英]Asp.net mvc return to controller from jsonresult
[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...
}
Instead of return Json(apptype, JsonRequestBehavior.AllowGet); 而不是返回Json(apptype,JsonRequestBehavior.AllowGet); how can i write below ?
我怎么在下面写?
return RedirectToAction("GetLawDetail", "Law", new { _lawID = baseappid });
If anybody wants to see Javascript: 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) {
}
....
..
How can i return to Law controller to GetLawDetail actionresult inside of jsonresult ? 如何在jsonresult内部将Law控制器返回到GetLawDetail actionresult ?
You can just return this action: 您可以返回以下操作:
return GetLawDetail(baseappid)
You should also change the return type to Action result as pointed by @im1dermike. 您还应按照@ im1dermike的指示将返回类型更改为“操作结果”。
Here's the complete code: 这是完整的代码:
[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...
}
Although, it will return you the whole rendered page. 虽然,它将返回您整个呈现的页面。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.