通过仅列表（空列表）从强类型列表单元格获取单元格DisplayName

Question

I have a list of "Book" (eg) as a model in view

@model List<Book>

I want to create a table that each column get it's header by Book's DisplayName propery:

<tr>
        <th class="text-right">
            @Html.DisplayNameFor(modelItem => Model.someMetaData.lineNumber)
        </th>
        <th class="text-right">
            @Html.DisplayNameFor(modelItem => Model.someMetaData.FirstName)
        </th>
<tr>

Answer 1

You can use DisplayNameFor helper method when your model is of type IEnumerable<T>

@model IEnumerable<Book>
<table class="table table-responsive table-striped">
    <tr>
        <th>@Html.DisplayNameFor(a=>a.Id)</th>
        <th>@Html.DisplayNameFor(a => a.Name)</th>
    </tr>

@foreach (var b in Model)
{
    <tr>
        <td>@b.Id</td>
        <td>@b.Name</td>
    </tr>
}
</table>

This works only if your view is strongly typed to an IEnumerable<Book> , It will not work for List<Book>