我想输出两个不同的视图(一个作为字符串，将作为电子邮件发送)，另一个是显示给用户的页面。
这在ASP中可能吗?NET MVC beta版?
我试过很多例子:
1. 在ASP。NET MVC Beta版
如果我使用这个例子，我会收到“HTTP后不能重定向”
报头已发送。”
2. MVC框架:捕获视图的输出
如果我使用这个，我似乎无法做一个重定向toaction，因为它
试图呈现一个可能不存在的视图。如果我返回视图
完全是一团糟，看起来一点都不对。
有人对我提出的这些问题有什么想法/解决方案吗?或者有什么更好的建议吗?
很多谢谢!
下面是一个例子。我要做的是创建GetViewForEmail方法:
public ActionResult OrderResult(string ref)
{
//Get the order
Order order = OrderService.GetOrder(ref);
//The email helper would do the meat and veg by getting the view as a string
//Pass the control name (OrderResultEmail) and the model (order)
string emailView = GetViewForEmail("OrderResultEmail", order);
//Email the order out
EmailHelper(order, emailView);
return View("OrderResult", order);
}
接受蒂姆·斯科特的回答(我做了一些修改和格式化):
public virtual string RenderViewToString(
ControllerContext controllerContext,
string viewPath,
string masterPath,
ViewDataDictionary viewData,
TempDataDictionary tempData)
{
Stream filter = null;
ViewPage viewPage = new ViewPage();
//Right, create our view
viewPage.ViewContext = new ViewContext(controllerContext, new WebFormView(viewPath, masterPath), viewData, tempData);
//Get the response context, flush it and get the response filter.
var response = viewPage.ViewContext.HttpContext.Response;
response.Flush();
var oldFilter = response.Filter;
try
{
//Put a new filter into the response
filter = new MemoryStream();
response.Filter = filter;
//Now render the view into the memorystream and flush the response
viewPage.ViewContext.View.Render(viewPage.ViewContext, viewPage.ViewContext.HttpContext.Response.Output);
response.Flush();
//Now read the rendered view.
filter.Position = 0;
var reader = new StreamReader(filter, response.ContentEncoding);
return reader.ReadToEnd();
}
finally
{
//Clean up.
if (filter != null)
{
filter.Dispose();
}
//Now replace the response filter
response.Filter = oldFilter;
}
}
示例使用
假设控制器通过Site调用订单确认电子邮件。主人的位置。
string myString = RenderViewToString(this.ControllerContext, "~/Views/Order/OrderResultEmail.aspx", "~/Views/Shared/Site.Master", this.ViewData, this.TempData);
下面是我为ASP编写的一个类。NETCore RC2。我使用它，所以我可以生成html电子邮件使用Razor。
using Microsoft.AspNetCore.Http;
using Microsoft.AspNetCore.Mvc;
using Microsoft.AspNetCore.Mvc.Abstractions;
using Microsoft.AspNetCore.Mvc.ModelBinding;
using Microsoft.AspNetCore.Mvc.Rendering;
using Microsoft.AspNetCore.Mvc.ViewEngines;
using Microsoft.AspNetCore.Mvc.ViewFeatures;
using Microsoft.AspNetCore.Routing;
using System.IO;
using System.Threading.Tasks;
namespace cloudscribe.Web.Common.Razor
{
/// <summary>
/// the goal of this class is to provide an easy way to produce an html string using
/// Razor templates and models, for use in generating html email.
/// </summary>
public class ViewRenderer
{
public ViewRenderer(
ICompositeViewEngine viewEngine,
ITempDataProvider tempDataProvider,
IHttpContextAccessor contextAccesor)
{
this.viewEngine = viewEngine;
this.tempDataProvider = tempDataProvider;
this.contextAccesor = contextAccesor;
}
private ICompositeViewEngine viewEngine;
private ITempDataProvider tempDataProvider;
private IHttpContextAccessor contextAccesor;
public async Task<string> RenderViewAsString<TModel>(string viewName, TModel model)
{
var viewData = new ViewDataDictionary<TModel>(
metadataProvider: new EmptyModelMetadataProvider(),
modelState: new ModelStateDictionary())
{
Model = model
};
var actionContext = new ActionContext(contextAccesor.HttpContext, new RouteData(), new ActionDescriptor());
var tempData = new TempDataDictionary(contextAccesor.HttpContext, tempDataProvider);
using (StringWriter output = new StringWriter())
{
ViewEngineResult viewResult = viewEngine.FindView(actionContext, viewName, true);
ViewContext viewContext = new ViewContext(
actionContext,
viewResult.View,
viewData,
tempData,
output,
new HtmlHelperOptions()
);
await viewResult.View.RenderAsync(viewContext);
return output.GetStringBuilder().ToString();
}
}
}
}
这个答案不在我的路上。这最初来自https://stackoverflow.com/a/2759898/2318354，但在这里我已经展示了如何使用它与“静态”关键字，使它通用于所有控制器。
为此，你必须在类文件中创建静态类。(假设你的类文件名是Utils.cs)
这个例子是剃须刀。
Utils.cs
public static class RazorViewToString
{
public static string RenderRazorViewToString(this Controller controller, string viewName, object model)
{
controller.ViewData.Model = model;
using (var sw = new StringWriter())
{
var viewResult = ViewEngines.Engines.FindPartialView(controller.ControllerContext, viewName);
var viewContext = new ViewContext(controller.ControllerContext, viewResult.View, controller.ViewData, controller.TempData, sw);
viewResult.View.Render(viewContext, sw);
viewResult.ViewEngine.ReleaseView(controller.ControllerContext, viewResult.View);
return sw.GetStringBuilder().ToString();
}
}
}
现在你可以从你的控制器调用这个类，通过在你的控制器文件中添加命名空间，通过将“this”作为参数传递给控制器，如下所示。
string result = RazorViewToString.RenderRazorViewToString(this ,"ViewName", model);
正如@Sergey给出的建议，这个扩展方法也可以从控制器调用，如下所示
string result = this.RenderRazorViewToString("ViewName", model);
我希望这将有助于您使代码干净整洁。
