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I have a table that I've built using a foreach loop in MVC 4 (Razor), and I'm trying to convert it to a WebGrid to take advantage of sorting. Currently, I'm just trying to get the format parameter to work for me so that I can recreate the columns as I need them. Here is a simple example that I can't get to work:

grid.Column(
    columnName: "ChangeId",
    format: (item) => @Html.DisplayFor(m => (item as ChangeStateViewModel).ChangeId)
)

This outputs nothing during rendering:

<td></td>

What am I doing wrong? I've tried removing (item) => and using <text>...</text>, but that doesn't display my value either.

EDIT

If I change it to the following, as mentioned here, it does display all of the properties in the item; however, building from there just puts me right back where I was.

format: @<text>@Html.DisplayFor(x => item)</text>

EDIT2

format: @<text>@item.ChangeId</text>

works, of course, but doesn't give me the ability to provide a DisplayTemplate.

EDIT3

It looks like (item as ChangeStateViewModel).ChangeId is giving me NULL.. but why? It's valued if I just use @item.ChangeId.

EDIT4

It looks like the default item provided (when you use @<text>...</text>) is a System.Web.Helpers.WebGridRow, which is why EDIT3 provides the behavior that it does. Something like the below will work, but only if you send it to a DisplayTemplate; for some reason, it won't show if you remove that , "DisplayAny" part.

format: @<text>@Html.DisplayFor(x => (item as WebGridRow)["ChangeId"], "DisplayAny")</text>

That does make EDIT2 rather confusing, though.

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zimdanen
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1 Answers1

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This works, so I'm stopping here and marking this as an answer. However, if there are better ways to approach this, I'd love to hear them.

format: @<text>@Html.DisplayFor(x => (item as WebGridRow)["ChangeId"], "DisplayAny")</text>

The DisplayTemplate is required; otherwise, this won't show.

zimdanen
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