Question

有没有办法向Action和Model组合发出GET请求以返回HTML，而不是向URI发出GET请求？

通常在我的代码中我会做

 return View("myView", myModel);

我想从中返回HTML，并且能够阅读HTML。

类似的东西：

 HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url);

但要传递动作和模型

Answer 1

有一些available here，您可以将模型作为参数传递。虽然仍然不确定为什么要这样做......替代来源available on github。

无论如何，如果您只是想以某种方式改变HTML，另一种方法是重定向到视图，然后使用动作过滤器，您可以在其中进行任何可能需要的后处理。

这样做会将任何视图渲染为HTML并将其作为String返回。虽然是较大类的一部分，但相关部分位于下方（Context实际为Controller.ControllerContext）：

/// <summary>
/// Internal method that handles rendering of either partial or 
/// or full views.
/// </summary>
/// <param name="viewPath">
/// The path to the view to render. Either in same controller, shared by 
/// name or as fully qualified ~/ path including extension
/// </param>
/// <param name="model">Model to render the view with</param>
/// <param name="partial">Determines whether to render a full or partial view</param>
/// <returns>String of the rendered view</returns>
protected string RenderViewToStringInternal(string viewPath, object model,
                                            bool partial = false)
{
    // first find the ViewEngine for this view
    ViewEngineResult viewEngineResult = null;
    if (partial)
        viewEngineResult = ViewEngines.Engines.FindPartialView(Context, viewPath);
    else
        viewEngineResult = ViewEngines.Engines.FindView(Context, viewPath, null);

    if (viewEngineResult == null)
        throw new FileNotFoundException(Resources.ViewCouldNotBeFound);

    // get the view and attach the model to view data
    var view = viewEngineResult.View;
    Context.Controller.ViewData.Model = model;

    string result = null;

    using (var sw = new StringWriter())
    {
        var ctx = new ViewContext(Context, view,
                                    Context.Controller.ViewData,
                                    Context.Controller.TempData,
                                    sw);
        view.Render(ctx, sw);
        result = sw.ToString();
    }

    return result;
}

从action / model返回HTML

1 个答案: