el modo de hacer la vista parcial en una cadena

Question

Tengo el siguiente código:

public ActionResult SomeAction() 
{ 
    return new JsonpResult 
    { 
     Data = new { Widget = "some partial html for the widget" } 
    }; 
} 

me gustaría modificarlo para que yo pudiera tener

public ActionResult SomeAction() 
{ 
    // will render HTML that I can pass to the JSONP result to return. 
    var partial = RenderPartial(viewModel); 
    return new JsonpResult 
    { 
     Data = new { Widget = partial } 
    }; 
} 

es esto posible? ¿Alguien podría explicar cómo?

nota, edité la pregunta antes de publicar la solución.

Answer 1

Ésta es una versión ligeramente modificada de una respuesta que funciona:

public static string RenderPartialToString(string controlName, object viewData) 
{ 
    ViewPage viewPage = new ViewPage() { ViewContext = new ViewContext() }; 

    viewPage.ViewData = new ViewDataDictionary(viewData); 
    viewPage.Controls.Add(viewPage.LoadControl(controlName)); 

    StringBuilder sb = new StringBuilder(); 
    using (StringWriter sw = new StringWriter(sb)) 
    { 
     using (HtmlTextWriter tw = new HtmlTextWriter(sw)) 
     { 
      viewPage.RenderControl(tw); 
     } 
    } 

    return sb.ToString(); 
} 

Uso:

string ret = RenderPartialToString("~/Views/MyController/MyPartial.ascx", model); 

Answer 2

de Dave,

una variación sobre el mismo tema (v1.0 MVC) :

 protected static string RenderPartialToString(Controller controller, string partialName, object model) 
    { 
     var vd = new ViewDataDictionary(controller.ViewData); 
     var vp = new ViewPage 
     { 
      ViewData = vd, 
      ViewContext = new ViewContext(), 
      Url = new UrlHelper(controller.ControllerContext.RequestContext) 
     }; 

     ViewEngineResult result = ViewEngines 
            .Engines 
            .FindPartialView(controller.ControllerContext, partialName); 

     if (result.View == null) 
     { 
      throw new InvalidOperationException(
      string.Format("The partial view '{0}' could not be found", partialName)); 
     } 
     var partialPath = ((WebFormView)result.View).ViewPath; 

     vp.ViewData.Model = model; 

     Control control = vp.LoadControl(partialPath); 
     vp.Controls.Add(control); 

     var sb = new StringBuilder(); 

     using (var sw = new StringWriter(sb)) 
     { 
      using (var tw = new HtmlTextWriter(sw)) 
      { 
       vp.RenderControl(tw); 
      } 
     } 
     return sb.ToString(); 
    } 

us edad dentro del controlador: Respuesta

 public string GetLocationHighlites() 
    { 
     IBlockData model = WebPagesMapper.GetLocationHighlites(); 
     // **this** being the controoler instance 
     // LocationPartial.ascx can be defined in shared or in view folder 
     return RenderPartialToString(**this**,"LocationPartial", model); 
    } 

Answer 3

de DaveDev funcionó bien para mí, sin embargo, cuando la vista parcial llama a otro parcial consigo "El valor no puede ser nulo. Nombre de parámetro:. vista"

Buscando en He hecho una variante del following que parece funcionar bien

public static string RenderPartialToString(string viewName, object model, ControllerContext ControllerContext) 
    { 
     if (string.IsNullOrEmpty(viewName)) 
      viewName = ControllerContext.RouteData.GetRequiredString("action"); 
     ViewDataDictionary ViewData = new ViewDataDictionary(); 
     TempDataDictionary TempData = new TempDataDictionary(); 
     ViewData.Model = model; 

     using (StringWriter sw = new StringWriter()) 
     { 
      ViewEngineResult viewResult = ViewEngines.Engines.FindPartialView(ControllerContext, viewName); 
      ViewContext viewContext = new ViewContext(ControllerContext, viewResult.View, ViewData, TempData, sw); 
      viewResult.View.Render(viewContext, sw); 

      return sw.GetStringBuilder().ToString(); 
     } 

    } 

Uso:

String result = MVCHelpers.RenderPartialToString("PartialViewHere", Model, ControllerContext) 

Answer 4

Puede crear extensión que hacen la vista en cadena

public static class RenderPartialToStringExtensions 
{ 
    /// <summary> 
    /// render PartialView and return string 
    /// </summary> 
    /// <param name="context"></param> 
    /// <param name="partialViewName"></param> 
    /// <param name="model"></param> 
    /// <returns></returns> 
    public static string RenderPartialToString(this ControllerContext context, string partialViewName, object model) 
    { 
     return RenderPartialToStringMethod(context, partialViewName, model); 
    } 

    /// <summary> 
    /// render PartialView and return string 
    /// </summary> 
    /// <param name="context"></param> 
    /// <param name="partialViewName"></param> 
    /// <param name="viewData"></param> 
    /// <param name="tempData"></param> 
    /// <returns></returns> 
    public static string RenderPartialToString(ControllerContext context, string partialViewName, ViewDataDictionary viewData, TempDataDictionary tempData) 
    { 
     return RenderPartialToStringMethod(context, partialViewName, viewData, tempData); 
    } 

    public static string RenderPartialToStringMethod(ControllerContext context, string partialViewName, ViewDataDictionary viewData, TempDataDictionary tempData) 
    { 
     ViewEngineResult result = ViewEngines.Engines.FindPartialView(context, partialViewName); 

     if (result.View != null) 
     { 
      StringBuilder sb = new StringBuilder(); 
      using (StringWriter sw = new StringWriter(sb)) 
      { 
       using (HtmlTextWriter output = new HtmlTextWriter(sw)) 
       { 
        ViewContext viewContext = new ViewContext(context, result.View, viewData, tempData, output); 
        result.View.Render(viewContext, output); 
       } 
      } 

      return sb.ToString(); 
     } 
     return String.Empty; 
    } 

    public static string RenderPartialToStringMethod(ControllerContext context, string partialViewName, object model) 
    { 
     ViewDataDictionary viewData = new ViewDataDictionary(model); 
     TempDataDictionary tempData = new TempDataDictionary(); 
     return RenderPartialToStringMethod(context, partialViewName, viewData, tempData); 
    } 
} 

Y luego úsala en acción

[HttpPost] 
public ActionResult GetTreeUnit(string id) 
{ 
    int _id = id.ExtractID(); 
    string render = ControllerContext.RenderPartialToString("SomeView"); 
    return Json(new { data = render }); 
} 

Answer 5

Opté por un método de extensión como el siguiente para una aplicación ASP.NET MVC 4. Creo que es más simple que algunas de las sugerencias que he visto:

public static class ViewExtensions 
{ 
    public static string RenderToString(this PartialViewResult partialView) 
    { 
     var httpContext = HttpContext.Current; 

     if (httpContext == null) 
     { 
      throw new NotSupportedException("An HTTP context is required to render the partial view to a string"); 
     } 

     var controllerName = httpContext.Request.RequestContext.RouteData.Values["controller"].ToString(); 

     var controller = (ControllerBase)ControllerBuilder.Current.GetControllerFactory().CreateController(httpContext.Request.RequestContext, controllerName); 

     var controllerContext = new ControllerContext(httpContext.Request.RequestContext, controller); 

     var view = ViewEngines.Engines.FindPartialView(controllerContext, partialView.ViewName).View; 

     var sb = new StringBuilder(); 

     using (var sw = new StringWriter(sb)) 
     { 
      using (var tw = new HtmlTextWriter(sw)) 
      { 
       view.Render(new ViewContext(controllerContext, view, partialView.ViewData, partialView.TempData, tw), tw); 
      } 
     } 

     return sb.ToString(); 
    } 
} 

Me permite hacer lo siguiente:

var html = PartialView("SomeView").RenderToString(); 

Además, este enfoque persiste cualquier Modelo, ViewBag y otros datos de vista para la vista.

Answer 6

funciona perfecto (Sólo ver el nombre requerido)
* para los parámetros que puede utilizar un modelo
* puede llamar a esto desde un punto de vista también

Vista lateral o lado llamante

BuyOnlineCartMaster ToInvoice1 = new BuyOnlineCartMaster(); // for passing parameters 
ToInvoice1.CartID = 1; 

string HtmlString = RenderPartialViewToString("PartialInvoiceCustomer", ToInvoice1); 

función de generación HTML

public static string RenderPartialViewToString(string viewName, object model) 
    {  
    using (var sw = new StringWriter()) 
      { 
       BuyOnlineController controller = new BuyOnlineController(); // instance of the required controller (you can pass this as a argument if needed) 

       // Create an MVC Controller Context 
       var wrapper = new HttpContextWrapper(System.Web.HttpContext.Current); 

       RouteData routeData = new RouteData(); 

       routeData.Values.Add("controller", controller.GetType().Name 
                  .ToLower() 
                  .Replace("controller", "")); 

       controller.ControllerContext = new ControllerContext(wrapper, routeData, controller); 

       controller.ViewData.Model = model; 

       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); 

       return sw.ToString(); 
      } 
    } 

Answer 7

Puede hacerlo de la caja con:

var partial = new HtmlString(Html.Partial("_myPartial", Model).ToString()); 

Answer 8

public virtual string RenderPartialViewToString(string viewName, object viewmodel) 
     { 
      if (string.IsNullOrEmpty(viewName)) 
      { 
       viewName = this.ControllerContext.RouteData.GetRequiredString("action"); 
      } 

      ViewData.Model = viewmodel; 

      using (var sw = new StringWriter()) 
      { 
       ViewEngineResult viewResult = System.Web.Mvc.ViewEngines.Engines.FindPartialView(this.ControllerContext, viewName); 
       var viewContext = new ViewContext(this.ControllerContext, viewResult.View, this.ViewData, this.TempData, sw); 
       viewResult.View.Render(viewContext, sw); 
       viewResult.ViewEngine.ReleaseView(ControllerContext, viewResult.View); 

       return sw.GetStringBuilder().ToString(); 
      } 
     }