ASP.NET视图组件：删除要查看的默认路径

Question

我想返回一个包含视图特定路径的视图，如return View(~/Views/Home)。要返回的类型是IViewComponentResult。

但是，当网站被渲染时，它会尝试在Components/{controller-name}/~/Views/Home下找到该视图。

所以我想问你们，如果你知道从路径中删除Components/{controller-name}/的聪明方法。我的视图组件类派生自ViewComponent类。

关于这一点的问题是，现在，你必须有一个“Components”文件夹，其控制器名称是一个包含Default.cshtml文件的子文件夹。我不喜欢将所有组件放在一个文件夹中。我希望你有一个解决方案。

Answer 1

约定发生在ViewViewComponentResult，因此如果您不想要约定，则必须实现自己的IViewComponentResult

由于mvc是开源的，你可以复制所有ViewViewComponentResult，然后更改约定位。

所以基本上要做两件事：

1. 创建自己的IViewComponentResult - 让我们称之为MyViewViewComponentResult
2. 在已实施的ViewComponent中创建帮助以替换原始View() - 目的是返回您在步骤1中创建的自定义MyViewViewComponentResult

创建自己的IViewComponentResult

惯例发生在ExecuteAsync：

public class ViewViewComponentResult : IViewComponentResult
{
    // {0} is the component name, {1} is the view name.
    private const string ViewPathFormat = "Components/{0}/{1}";
    private const string DefaultViewName = "Default";

    ....

    public async Task ExecuteAsync(ViewComponentContext context)
    {
        ....
        if (result == null || !result.Success)
        {
            // This will produce a string like:
            //
            //  Components/Cart/Default
            //
            // The view engine will combine this with other path info to search paths like:
            //
            //  Views/Shared/Components/Cart/Default.cshtml
            //  Views/Home/Components/Cart/Default.cshtml
            //  Areas/Blog/Views/Shared/Components/Cart/Default.cshtml
            //
            // This supports a controller or area providing an override for component views.
            var viewName = isNullOrEmptyViewName ? DefaultViewName : ViewName;
            var qualifiedViewName = string.Format(
                CultureInfo.InvariantCulture,
                ViewPathFormat,
                context.ViewComponentDescriptor.ShortName,
                viewName);

            result = viewEngine.FindView(viewContext, qualifiedViewName, isMainPage: false);
        }

        ....
    }

    .....
}

所以根据需要改变它

在已实施的ViewComponent中创建一个帮助器，替换原来的View

基于原始

/// <summary>
/// Returns a result which will render the partial view with name <paramref name="viewName"/>.
/// </summary>
/// <param name="viewName">The name of the partial view to render.</param>
/// <param name="model">The model object for the view.</param>
/// <returns>A <see cref="ViewViewComponentResult"/>.</returns>
public ViewViewComponentResult View<TModel>(string viewName, TModel model)
{
    var viewData = new ViewDataDictionary<TModel>(ViewData, model);
    return new ViewViewComponentResult
    {
        ViewEngine = ViewEngine,
        ViewName = viewName,
        ViewData = viewData
    };
}

你会做类似

的事情

/// <summary>
/// Returns a result which will render the partial view with name <paramref name="viewName"/>.
/// </summary>
/// <param name="viewName">The name of the partial view to render.</param>
/// <param name="model">The model object for the view.</param>
/// <returns>A <see cref="ViewViewComponentResult"/>.</returns>
public MyViewViewComponentResult MyView<TModel>(string viewName, TModel model)
{
    var viewData = new ViewDataDictionary<TModel>(ViewData, model);
    return new MyViewViewComponentResult
    {
        ViewEngine = ViewEngine,
        ViewName = viewName,
        ViewData = viewData
    };
}

以后您可以拨打MyView()而不是View()

更多信息请查看另一个SO：Change component view location in Asp.Net 5