返回局部视图，如果动作是AJAX或儿童普遍的行动 [英] Return partial view if action is ajax or child action universally
问题描述
家伙，您好我已经烧起来了谷歌试图建立某种类的，这将决定的通用如果它是一个AJAX调用或儿童行为或行动，使我的控制器可以决定是否返回部分视图或全视图。到目前为止，我还没有多少运气。目前我使用下面的code来实现这个
Hi guys I have burned up google trying to construct some kind of class that will determine UNIVERSALLY if it is a AJAX call or child action or action so that my controllers can determine whether to return a partial view or a full view. So far I havent had much luck. At the moment I am using the following code to achieve this
if (Request.IsAjaxRequest() || ControllerContext.IsChildAction)
{
return PartialView();
}
return View();
问题是你必须这样做，在一个控制器和每一个条件，你遇到的每一个动作，但我相信有一种方法通过一个帮手来实现这一点，但找不出如何。可以请你点我的任何链接/例如code来实现这一目标。
The problem is you have to do this in every action in a controller and every Condition you encounter but I am sure there is a way to achieve this via a helper but cant figure out how. Can you please point me to any links/example code to achieve this.
编辑：
@Aron我已经发布了一块code作为整个控制器就太长了。但是你可以看到我的predicament。返回包含查看和对象/型号K。
@Aron I have posted a piece of code as whole controller would be too long. But you can see my predicament. The return contains a View and a Object/Model "k".
public ActionResult _Details_Message(int id = 0, int CId = 0)
{
ViewBag.MrnSortParm = CId;
if (id != 0)
{
var k = mrn.MRNS.Where(u => u.Id == id).SingleOrDefault();
if (k.To == User.Identity.Name)
{
if (k.Type == 0) // message
{
k.Read = true;
mrn.Entry(k).State = EntityState.Modified;
mrn.SaveChanges();
}
return PartialView("_Details_Message", k);//replace the above code here
}
if (k.From == User.Identity.Name)
{
return PartialView("_Sent", k); //replace the above code here
}
}
var m = new message();
m.CourierId = CId;
return PartialView("_Create_Message", m); //replace the above code here
}
编辑2
我已经发现一个答案它不是一个辅助功能，但在该视图的变形例。链接是<一个href=\"http://stackoverflow.com/questions/5318385/mvc-3-how-to-render-a-view-without-its-layout-page\">here 。可以标记我自己的问题是重复的：（
Edit 2 I have found an answer it isn't an helper function but a modification in the view. The link is here . Can mark my own question as duplicate :(
推荐答案
那么你很幸运，因为我写了code一吨做同样的事情。如果你想在模型返回一个JSON对象或查看这也考虑到了。它还包装所有Ajax调用变成一个包装反应元件
Well you're in luck because I wrote a TON of code to do something similar. This also takes into account if you want to return the model as a JSON object or a View. It also wraps all the Ajax calls into a wrapper response element
基本上，如果你有一个UI的家伙做的东西，你永远不需要知道他想要什么。让他写的意见，或进行AJAX调用。这完全解耦从C＃开发的UI人（只要他懂得如何编写MVC的看法，他并不需要在所有的控制器是如何工作的，传递的只是模型知道）。
Basically if you have a UI guy doing stuff, you NEVER need to know what he wants. Let him write the views, or make AJAX calls. This completely decouples the UI person from the C# developer (as long as he understands how to write MVC Views, he doesn't need to know at all how the controller works, just the model being passed).
的 ControllerBase 类：
public abstract class MyControllerBase : Controller
{
// could be moved to web.config
private const _jsonDataType = "JsonDataType";
public bool IsAjaxRequest
{
get
{
return this.HttpContext.Request.IsAjaxRequest();
}
}
public bool IsAjaxHtmlRequest
{
get
{
return string.Equals(this.Request.Headers[MyControllerBase._jsonDataType], "html", StringComparison.CurrentCultureIgnoreCase);
}
}
private JsonResponse GetAjaxResponse()
{
JsonResponse result = new JsonResponse();
result.IsValid = true;
return result;
}
private JsonResponse<T> GetAjaxResponse<T>(T model)
{
JsonResponse<T> result = new JsonResponse<T>();
result.Data = model;
result.IsValid = true;
return result;
}
private JsonResponse<string> GetAjaxHtmlResponse()
{
JsonResponse<string> result = new JsonResponse<string>();
result.Data = this.PartialViewToString(this.ControllerContext.RouteData.Values["Action"].ToString(), null);
result.IsValid = true;
return result;
}
private JsonResponse<string> GetAjaxHtmlResponse<T>(T model)
{
JsonResponse<string> result = new JsonResponse<string>();
result.Data = this.PartialViewToString(this.ControllerContext.RouteData.Values["Action"].ToString(), model);
result.IsValid = true;
return result;
}
private JsonResponse<string> GetAjaxHtmlResponse<T>(T model, string viewName)
{
JsonResponse<string> result = new JsonResponse<string>();
result.Data = this.PartialViewToString(viewName, model);
result.IsValid = true;
return result;
}
public ActionResult ViewOrAjax()
{
return this.ViewOrAjax(JsonRequestBehavior.DenyGet);
}
public ActionResult ViewOrAjax(JsonRequestBehavior jsonRequestBehavior)
{
if (this.ControllerContext.IsChildAction)
{
return this.PartialView(this.ControllerContext.RouteData.Values["Action"].ToString(), null);
}
if (this.IsAjaxRequest)
{
if (this.IsAjaxHtmlRequest)
{
return this.Json(this.GetAjaxHtmlResponse(), jsonRequestBehavior);
}
return this.Json(this.GetAjaxResponse(), jsonRequestBehavior);
}
return this.View(this.ControllerContext.RouteData.Values["Action"].ToString(), null);
}
public ActionResult ViewOrAjax<T>(T model)
{
return this.ViewOrAjax<T>(model, JsonRequestBehavior.DenyGet);
}
public ActionResult ViewOrAjax<T>(T model, JsonRequestBehavior jsonRequestBehavior)
{
if (this.ControllerContext.IsChildAction)
{
return this.PartialView(model);
}
if (this.IsAjaxRequest)
{
if (this.IsAjaxHtmlRequest)
{
return this.Json(this.GetAjaxHtmlResponse(model), jsonRequestBehavior);
}
return this.Json(this.GetAjaxResponse<T>(model), jsonRequestBehavior);
}
return this.View(model);
}
public ActionResult ViewOrAjax<T>(IView view, T model, JsonRequestBehavior jsonRequestBehavior)
{
if (this.ControllerContext.IsChildAction)
{
return this.PartialView(model);
}
if (this.IsAjaxRequest)
{
if (this.IsAjaxHtmlRequest)
{
return this.Json(this.GetAjaxHtmlResponse(model), jsonRequestBehavior);
}
return this.Json(this.GetAjaxResponse<T>(model), jsonRequestBehavior);
}
return this.View(view, model);
}
public ActionResult ViewOrAjax<T>(string viewName, T model)
{
return this.ViewOrAjax<T>(viewName, model, JsonRequestBehavior.DenyGet);
}
public ActionResult ViewOrAjax<T>(string viewName, T model, JsonRequestBehavior jsonRequestBehavior)
{
if (this.ControllerContext.IsChildAction)
{
return this.PartialView(model);
}
if (this.IsAjaxRequest)
{
if (this.IsAjaxHtmlRequest)
{
return this.Json(this.GetAjaxHtmlResponse(model, viewName), jsonRequestBehavior);
}
return this.Json(this.GetAjaxResponse<T>(model), jsonRequestBehavior);
}
return this.View(viewName, model);
}
public ActionResult ViewOrAjax<T>(string viewName, string masterName, T model)
{
return this.ViewOrAjax<T>(viewName, masterName, model, JsonRequestBehavior.DenyGet);
}
public ActionResult ViewOrAjax<T>(string viewName, string masterName, T model, JsonRequestBehavior jsonRequestBehavior)
{
if (this.ControllerContext.IsChildAction)
{
return this.PartialView(model);
}
if (this.IsAjaxRequest)
{
if (this.IsAjaxHtmlRequest)
{
return this.Json(this.GetAjaxHtmlResponse(model, viewName), jsonRequestBehavior);
}
return this.Json(this.GetAjaxResponse(model), jsonRequestBehavior);
}
return this.View(viewName, masterName, model);
}
protected internal new ViewResult View(string viewName, string masterName, object model)
{
if (model != null)
{
ViewData.Model = model;
}
ViewResult result = new ViewResult
{
ViewName = viewName,
MasterName = masterName,
ViewData = ViewData,
TempData = TempData
};
return result;
}
}
的 JsonResponse＆LT;＆GT; 阿贾克斯全球包装呼叫：
The JsonResponse<> global wrapper for Ajax Calls:
public class JsonResponse
{
public JsonResponse()
{
}
public bool IsValid { get; set; }
public bool IsAjaxRequestUnsupported { get; set; }
public string RedirectTo { get; set; }
public string CanonicalUrl { get; set; }
}
public class JsonResponse<T> : JsonResponse
{
public JsonResponse() : base()
{
}
public T Data { get; set; }
}
JavaScript的 global_getJsonResponse code（需要jQuery的）：
The Javascript global_getJsonResponse code (require jQuery):
function global_getJsonResult(Controller, View, data, successCallback, completeCallback, methodType, returnType, jsonDataType) {
if (IsString(Controller)
&& IsString(View)
&& !IsUndefinedOrNull(data)) {
var ajaxData;
var ajaxType;
if (typeof (data) == "string") {
ajaxData = data;
ajaxType = "application/x-www-form-urlencoded"
}
else {
ajaxData = JSON.stringify(data);
ajaxType = "application/json; charset=utf-8";
}
var method = 'POST';
if (methodType) {
method = methodType;
}
var dataType = 'json';
if (returnType) {
dataType = returnType;
}
var jsonType = 'html';
if (jsonDataType) {
jsonType = jsonDataType;
}
var jqXHR = $.ajax({
url: '/' + Controller + '/' + View,
headers: { JsonDataType: jsonType },
data: ajaxData,
type: method,
dataType: dataType,
contentType: ajaxType,
success: function (jsonResult) {
if (!IsUndefinedOrNull(jsonResult)
&& jsonResult.hasOwnProperty("RedirectTo")
&& !IsUndefinedOrNull(jsonResult.RedirectTo)
&& jsonResult.RedirectTo.length > 0) {
$.fn.notify('error', 'Login Expired', 'You have been inactive for a prolonged period of time, and have been logged out of the system.');
window.setTimeout(function () { window.location = jsonResult.RedirectTo }, 5000);
}
else if (IsFunction(successCallback)) {
successCallback(jsonResult, Controller + '/' + View);
}
},
error: function (jqXHR, textStatus, errorThrown) {
if (errorThrown != 'abort') {
$.fn.notify('error', 'Whoops! Something went wrong.', 'We have been notified of the error.'/* textStatus + ': ' + errorThrown*/);
}
log('ERROR IN global_getJsonResult() : ', textStatus, errorThrown, jqXHR);
},
complete: function (jqXHR, textStatus) {
if (IsFunction(completeCallback)) {
completeCallback(jqXHR, textStatus, Controller + '/' + View);
}
}
});
return jqXHR;
}
}
这code支持通过<一个包括服务器端和客户端超时href=\"http://stackoverflow.com/questions/5238854/handling-session-timeout-in-ajax-calls/5242746#5242746\">Handling在阿贾克斯会话超时调用，像一个变化：
This code supports both server side and client side TimeOuts via Handling session timeout in ajax calls, with a change like:
protected override void HandleUnauthorizedRequest(AuthorizationContext filterContext)
{
if (filterContext.HttpContext.Request.IsAjaxRequest())
{
filterContext.Result = new JsonResult
{
Data = new JsonResponse<bool>
{
IsValid = false,
RedirectTo = FormsAuthentication.LoginUrl
},
JsonRequestBehavior = JsonRequestBehavior.AllowGet
};
}
else
{
base.HandleUnauthorizedRequest(filterContext);
}
}
一对夫妇控制器扩展方法，让你返回在JSON文本（这code是从SO，我通常这样的文件，但我失去了它）呈现局部视图：
A couple of Extension Methods on controller to allow you to return rendered partial views as text in json (this code was from SO, I usually document such but I lost it):
internal static class ControllerExtensions
{
public static string PartialViewToString(this Controller instance, object model)
{
string viewName = instance.ControllerContext.RouteData.GetRequiredString("action");
return ControllerExtensions.PartialViewToString(instance, viewName, model);
}
public static string PartialViewToString(this Controller instance, string viewName, object model)
{
string result;
ViewDataDictionary viewData = instance.ViewData;
viewData.Model = model;
using (var sw = new StringWriter())
{
var viewResult = ViewEngines.Engines.FindPartialView(instance.ControllerContext, viewName);
var viewContext = new ViewContext(instance.ControllerContext, viewResult.View, viewData, instance.TempData, sw);
viewResult.View.Render(viewContext, sw);
viewResult.ViewEngine.ReleaseView(instance.ControllerContext, viewResult.View);
result = sw.GetStringBuilder().ToString();
}
return result;
}
}
现在得到（可惜）所有的控制器从这个基本控制器：
Now derive (sadly) all your controllers from this base controller:
public HomeController : MyBaseController
{
public ActionResult Index()
{
var viewModel = new MyViewModel();
return this.ViewOrAjax(viewModel);
}
}
现在如果页面是由浏览器作为标准被调用，你会得到一个正常的布局呈现在页面（又名 this.View（视图模型））
Now if the page is called by the browser as your standard get, you get the page rendered normally with a Layout (aka this.View(viewModel)).
如果你调用它通过JavaScript的使用Ajax：
If you call it using Ajax via the Javascript:
global_getJsonResult("Home", // Controller or 'Area/Home' for areas
"Index", // View
$('#form').serialize(), // Json object or a serialized Form
jsCallBack, // call back function or null
"Post", // Get or Post
"Html"); // "Html" to return a Partial View in "Data"
// or "Json" to return a serialized view model in "Data"
这篇关于返回局部视图，如果动作是AJAX或儿童普遍的行动的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
