Change component view location in Asp.Net 5

Question

On ASP.NET 5 a Component view must be in one of two places:

Views/NameOfControllerUsingComponent/Components/ComponentName/Default.cshtml
Views/Shared/Components/ComponentName/Default.cshtml

Is there a way to change this to:

Views/NameOfControllerUsingComponent/Components/ComponentName.cshtml
Views/Shared/Components/ComponentName.cshtml

So basically, remove the folder ComponentName and change the view name from Default.cshtml to ComponentName.cshtml.

For me it makes more sense ... Is it possible?

Answer 1

该约定仅适用于创建从ViewComponent框架提供的基础派生的视图组件。

该类定义了View帮助程序，它返回一个ViewViewComponentResult：

public ViewViewComponentResult View<TModel>(string viewName, TModel model)
{
    var viewData = new ViewDataDictionary<TModel>(ViewData, model);
    return new ViewViewComponentResult
    {
        ViewEngine = ViewEngine,
        ViewName = viewName,
        ViewData = viewData
    };
}

这ViewViewComponentResult是定义约定的地方：

private const string ViewPathFormat = "Components/{0}/{1}";
private const string DefaultViewName = "Default";

public async Task ExecuteAsync(ViewComponentContext context)
{
    ...

    string qualifiedViewName;
    if (!isNullOrEmptyViewName &&
        (ViewName[0] == '~' || ViewName[0] == '/'))
    {
        // View name that was passed in is already a rooted path, the view engine will handle this.
        qualifiedViewName = ViewName;
    }
    else
    {
        // 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;

        qualifiedViewName = string.Format(
            CultureInfo.InvariantCulture,
            ViewPathFormat,
            context.ViewComponentDescriptor.ShortName,
            viewName);
    }

    ...

}

请注意，如果您从视图组件返回视图的完整路径作为视图名称，则视图组件将使用指定的视图。就像是：

return View("~/Views/Shared/Components/ComponentName.cshtml")

由于无法修改中的约定，ViewViewComponentResult并且您的方法仅适用于具有单个视图的视图组件，因此您可以使用根视图路径方法构建一些内容：

• 创建您自己的ViewComponent类，扩展现有的类。

• 添加新的辅助方法或隐藏现有View方法以使用完整路径返回视图：

  public ViewViewComponentResult MyView<TModel>(TModel model)
  {
      var viewName = string.Format(
              "~/Views/Shared/Components/{0}.cshtml", 
              this.ViewComponentContext.ViewComponentDescriptor.ShortName)
      return View(viewName, model);
  }
  

• 如果您添加新方法，您可以将它们添加为扩展方法，ViewComponent而不必创建自己的类。

另一种选择是创建一个类来SingleViewViewComponent复制代码ViewComponent但替换ViewViewComponentResult View<TModel>(string viewName, TModel model). 然后在创建视图组件时，您将继承 fromSingleViewViewComponent而不是ViewComponent.