How to Return partial view of another controller by controller?

I have an XXX.cshtml file in a Views\ABC folder. Its controller is ABC

I also have an action method in my DEF controller that return a Partialview("XXX" , xyzmodel)

I get a "view not found" error.

How to call that view from other controller

Solution 1:

Normally the views belong with a specific matching controller that supports its data requirements, or the view belongs in the Views/Shared folder if shared between controllers (hence the name).

"Answer" (but not recommended - see below):

You can refer to views/partial views from another controller, by specifying the full path (including extension) like:

return PartialView("~/views/ABC/XXX.cshtml", zyxmodel);

or a relative path (no extension), based on the answer by @Max Toro

return PartialView("../ABC/XXX", zyxmodel);

BUT THIS IS NOT A GOOD IDEA ANYWAY

*Note: These are the only two syntax that work. not ABC\\XXX or ABC/XXX or any other variation as those are all relative paths and do not find a match.

Better Alternatives:

You can use Html.Renderpartial in your view instead, but it requires the extension as well:

Html.RenderPartial("~/Views/ControllerName/ViewName.cshtml", modeldata);

Use @Html.Partial for inline Razor syntax:

@Html.Partial("~/Views/ControllerName/ViewName.cshtml", modeldata)

You can use the ../controller/view syntax with no extension (again credit to @Max Toro):

@Html.Partial("../ControllerName/ViewName", modeldata)

Note: Apparently RenderPartial is slightly faster than Partial, but that is not important.

If you want to actually call the other controller, use:

@Html.Action("action", "controller", parameters)

Recommended solution: @Html.Action

My personal preference is to use @Html.Action as it allows each controller to manage its own views, rather than cross-referencing views from other controllers (which leads to a large spaghetti-like mess).

You would normally pass just the required key values (like any other view) e.g. for your example:

@Html.Action("XXX", "ABC", new {id = model.xyzId })

This will execute the ABC.XXX action and render the result in-place. This allows the views and controllers to remain separately self-contained (i.e. reusable).

Update Sep 2014:

I have just hit a situation where I could not use @Html.Action, but needed to create a view path based on a action and controller names. To that end I added this simple View extension method to UrlHelper so you can say return PartialView(Url.View("actionName", "controllerName"), modelData):

public static class UrlHelperExtension
{
    /// <summary>
    /// Return a view path based on an action name and controller name
    /// </summary>
    /// <param name="url">Context for extension method</param>
    /// <param name="action">Action name</param>
    /// <param name="controller">Controller name</param>
    /// <returns>A string in the form "~/views/{controller}/{action}.cshtml</returns>
    public static string View(this UrlHelper url, string action, string controller)
    {
        return string.Format("~/Views/{1}/{0}.cshtml", action, controller);
    }
}

Solution 2:

The control searches for a view in the following order:

• First in shared folder
• Then in the folder matching the current controller (in your case it's Views/DEF)

As you do not have xxx.cshtml in those locations, it returns a "view not found" error.

Solution: You can use the complete path of your view:

Like

 PartialView("~/views/ABC/XXX.cshtml", zyxmodel);

Solution 3:

Simply you could use:

PartialView("../ABC/XXX")