ASP.NET MVC partial view and form action name

Question

How do I create a partial view that has a form with assigned id? I got as far as:

using (Html.BeginForm(?action?,"Candidate",FormMethod.Post,new {id="blah"}))

Partial view is used for both Create and Edit so first parameter ?action? will be different. I can't figure out what value of ?action? supposed to be.

UPDATE:

I guess I was not clear enough with the question. What I ended up doing is splitting Request.RawUrl to get controller name and action name:

 string[] actionUrlParts = ViewContext.HttpContext.Request.RawUrl.Split('/');
 using (Html.BeginForm(actionUrlParts.Length >= 2? actionUrlParts[2] : "",
        actionUrlParts.Length >= 1 ? actionUrlParts[1] : "", FormMethod.Post, new { id = "blah" }))

Kind of ugly but it works. Is there a better way to get an action name inside the partial view?

tvanfosson · Accepted Answer · 2009-04-09 14:25:55Z

7

Pass in the action to be performed via ViewData.

In your action that renders the view, create a ViewData item for the postback action. In your form reference this ViewData item to fill in the action parameter. Alternatively, you can create a view-only model that includes the action and the actual model as properties and reference it from there.

Example using ViewData:

using (Html.BeginForm( (string)ViewData["PostBackAction"], "Candidate", ...

Rendering actions:

public ActionResult Create()
{
     ViewData["PostBackAction"] = "New";
     ...
}


public ActionResult Edit( int id )
{
     ViewData["PostBackAction'] = "Update";
     ...
}

Example using Model

public class UpdateModel
{
     public string Action {get; set;}
     public Candidate CandidateModel { get; set; }
}

using (Html.BeginForm( Model.Action, "Candidate", ...

Rendering actions:

public ActionResult Create()
{
     var model = new UpdateModel { Action = "New" };

     ...

     return View(model);
}


public ActionResult Edit( int id )
{
     var model = new UpdateModel { Action = "Update" };

     model.CandidateModel = ...find corresponding model from id...

     return View(model);
}

EDIT: Based on your comment, if you feel that this should be done in the view (though I disagree), you could try some logic based off the ViewContext.RouteData

<%
    var action = "Create";
    if (this.ViewContext.RouteData.Values["action"] == "Edit")
    {
        action = "Update";
    }
    using (Html.BeginForm( action, "Candidate", ... 
    {
 %>

edited Apr 9, 2009 at 14:25

answered Apr 8, 2009 at 20:08

tvanfosson

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5 Comments

Dmitriy Shvadskiy Over a year ago

This will work but I don't really want to change Controller code to handle something that View should be able to figure out

tvanfosson Over a year ago

I added alternative this is entirely view-based, though I would think that the proper place to control what action to postback to would be in the controller.

Dmitriy Shvadskiy Over a year ago

Yep. That's exactly what I was looking for. The action name is in ViewContext.RouteData.Values["action"]

tvanfosson Over a year ago

@Dmitriy -- updated to use values property in example. I still think that the controller ought to control the flow in the app either by choosing a view with a fixed postback url or supplying the postback url to a view with a dynamic one. Views should responsible for rendering, not control logic.

Norman Over a year ago

Thank you for the info as it solved an issue I had with extending some conditional functionality to a Partial. Re the code example of using the ViewContext.RouteData I needed to cast it as a string for the comparision as below but was great for getting me in the right direciton. <%if ((string)this.ViewContext.RouteData.Values["action"] == "Create") {%> - Thanks, Norman

rodpl · Accepted Answer · 2010-03-21 19:26:14Z

1

Pass nulls as action and controller. Extension will use just current action and current controller

using (Html.BeginForm(null, null, FormMethod.Post, new { id="Model" }))

Action generated for form will be the same as parent view of this partial view.

It generates

<form action="/Orders/Edit/1" id="Model" method="post">

for url http://localhost:1214/Orders/Edit/1

... and this

<form action="/Orders/Create" id="Model" method="post">

for url http://localhost:1214/Orders/Create

answered Mar 21, 2010 at 19:26

rodpl

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Comments

johnnyRose · Accepted Answer · 2017-09-08 18:10:10Z

0

<% html.RenderPartial("MyUserControl", Model.ID) %>

edited Sep 8, 2017 at 18:10

johnnyRose

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answered Apr 8, 2009 at 20:16

Al Katawazi

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2 Comments

Will Barnwell Over a year ago

Hello, please do not post code only answers, your answer could be greatly improved with some explanation of what your code is doing and how this solves OP's problem.

Eric Leschinski Over a year ago

Code only answers are low quality and subject to deletion, there should always be at least one or two sentences (optimally a paragraph) about what question you are addressing assumptions, conditions, pitfalls, etc).

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