using surfacecontroller

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Nhan Pham 14 posts 84 karma points

Apr 21, 2016 @ 08:57

0

Using SurfaceController

Using Umbraco And Getting Started

Umbraco 6

I create partial view and using surfacecontroller.

But I want to post model back to View after post function. I can't return model back to view.

My code:

Controller:

View:

Before POST:

Input data

After POST:

After POST data not save. It will reset.

I already use "CurrentUmbracoPage"

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Dennis Adolfi 1082 posts 6446 karma points MVP 5x c-trib

Apr 21, 2016 @ 09:36
0
Hi Nhan!

Try:
```
return RedirectToCurrentUmbracoPage();
```
About return CurrentUmbracoPage() (By Markus Knappen Johansson):

Generally you want to return to the current page if you want to keep the POST-data that was sent to the controller.

About return RedirectToCurrentUmbracoPage() (By Shannon Deminick.):

If your POST is successful, you shouldn't just return a response, you should redirect to a page so you can avoid the aweful issue of having a user press refresh and have it resubmit the POST

http://www.enkelmedia.se/blogg/2014/5/8/currentumbracopage-or-redirecttocurrentumbracopage.aspx

Best of luck to you, let me know if it works out for you!
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Bo Damgaard Mortensen 719 posts 1207 karma points

Apr 21, 2016 @ 09:42

0

In addition to Dennis's post: also consider: return RedirectToCurrentUmbracoUrl(); if you're passing in querystring parameters for your page, these will be preserved.

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Nhan Pham 14 posts 84 karma points

Apr 21, 2016 @ 09:51

0

Hi Denis.

I want to return model to View back. I arealdy using return CurrentUmbracoPage(); If using return RedirectToCurrentUmbracoPage(); my model data will be lost.

But I use return CurrentUmbracoPage(); data from model cannot pass to View.

Please help me

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Dennis Adolfi 1082 posts 6446 karma points MVP 5x c-trib

Apr 21, 2016 @ 10:01

0

Since you have the form in a partialview, i dont think you can edit the model that´s beeing passed in and then return it to the currentpage, since your currentpage´s model is not CheckCodeModel, but probobly of type UmbracoTeplatePage or UmbracoViewPage.

If you where to re-buildthe form to a Ajax formpost you could return the model CheckCodeModel to the same partial view and not having to do a postback. Read more about Ajax formposts with Umbraco here: http://www.systenics.com/blog/creating-an-ajax-enabled-contact-form-in-umbraco-6-with-aspnet-mvc-4-and-twitter-bootstrap/?tag=Umbraco+6

Another way would be to pass the values back to the view in a TempData object instead of ViewBags, since Tempdata objects "survives" the redirect.

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Nhan Pham 14 posts 84 karma points

Apr 21, 2016 @ 10:20

0

Can I do that without partial view?

Write in template page?

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Dennis Adolfi 1082 posts 6446 karma points MVP 5x c-trib

Apr 21, 2016 @ 10:08

Controller:

    [ChildActionOnly]
    public ActionResult Index()
    {
    var code = (string)(TempData["code"] ?? "sjfkldsf");
    return PartialView("_CheckCode", new CheckCodeModel() { Code = code });
    }

[HttpPost]
[NotChildAction]
public ActionResult Index(CheckCodeModel model)
{
TempData["code"] = Model.Code;
TempData["name"] = "djhfkjdfkdjfkdjf";
return RedirectToCurrentUmbracoPage();
}

View:

<div>@Html.Raw(Model != null ? Model.Code : "")</div>
    <div>@Html.Raw(TempData["name"])</div>

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Nhan Pham 14 posts 84 karma points

Apr 21, 2016 @ 10:26

0

Thank Denis!

But I want pass ModelState to View, etc.

I will try another solution.

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Dennis Adolfi 1082 posts 6446 karma points MVP 5x c-trib

Apr 21, 2016 @ 10:57

0

I see. Then look at doing it with Ajax Form, that will work with ModelState.

Good luck to you!

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