3

尝试使用示例http://weblog.west-wind.com/posts/2012/May/30/Rendering-ASPNET-MVC-Views-to-String将 VIEW 作为字符串传递，并作为电子邮件发送。哪个应该通过电子邮件将销售发票发送给用户。

我已将 ViewRenderer 类添加到我的项目中。然后将 ContactSeller 函数添加到我的控制器中，并将发票视图复制并重命名为ViewOrderThroughEmail.cshtml

  [HttpPost]
    [AlwaysAccessible]
    public ActionResult SendEmailAttachment(QBCustomerRecord cust)
    {
        ContactSellerViewModel model = new ContactSellerViewModel();
        string invoiceEmailAsString = ContactSeller(model);
        _userService.SendEmail(username, nonce => Url.MakeAbsolute(Url.Action("LostPassword", "Account", new { Area = "Orchard.Users", nonce = nonce }), siteUrl), invoiceEmailAsString);

        _orchardServices.Notifier.Information(T("The user will receive a confirmation link through email."));

        return RedirectToAction("LogOn");
    }

    [HttpPost]
    public string ContactSeller(ContactSellerViewModel model)
    {   
        string message = ViewRenderer.RenderView("~/Orchard.Web/Modules/RainBow/Views/Account/ViewOrderThroughEmail.cshtml",model,
                                                     ControllerContext);     
        model.EntryId = 101;
        model.EntryTitle = message;


        return message;
    }

但这会引发错误，VIEW 不能为 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();
    }

在RenderViewToStringInternalViewRenderer.cs 的函数中。我原本以为这是视图的路径，但事实并非如此。

有任何想法吗？谢谢

4

1 回答 1

2

我在我的项目中使用以下扩展方法：

public static string RenderView(this Controller controller, string viewName, ViewDataDictionary viewData)
{
    var controllerContext = controller.ControllerContext;

    var viewResult = ViewEngines.Engines.FindView(controllerContext, viewName, null);

    StringWriter stringWriter;

    using (stringWriter = new StringWriter())
    {
        var viewContext = new ViewContext(
            controllerContext,
            viewResult.View,
            viewData,
            controllerContext.Controller.TempData,
            stringWriter);

        viewResult.View.Render(viewContext, stringWriter);
        viewResult.ViewEngine.ReleaseView(controllerContext, viewResult.View);
    }

    return stringWriter.ToString();
}

public static string RenderView(this Controller controller, string viewName, object model)
{
    return RenderView(controller, viewName, new ViewDataDictionary(model));
}

然后在您的操作方法中：

var viewString = this.RenderView("ViewName", model);
于 2014-01-07T10:49:58.923 回答