I have a .net core MVC project that uses AJAX to retrieve data when loading grids.
Today, both the function returning the view and the function returning the data for the grid, is in the same controller. This is not optimal for many reasons. I.e. I would like a Json ProblemResult to be returned if a exception occurs when calling function using AJAX, but when returning a View, I would like the Developer Exception page to be shown if an error occurs.
I could split the functions into different Controllers and annotate one of them with the ApiController attribute, but since the project has several hundreds of controllers it would be a significant task to do so.
What I would like is this:
If context type is application/json: Do model validation and return a “ProblemResult” if a exception occurs, otherwise use the Developers Exception Page to show the error.
Can this be done in a easy way, or do I need to build a middleware and handle it all by myself?
The easy way, meaning, doing this from a function that returns either a view or a type of JSON result would look something like this:
public IActionResult AjaxOrView(CheckModel model)
{
var isAjax = Request.Headers["X-Requested-With"] == "XMLHttpRequest";
var modelStateValid = ModelState.IsValid;
if (isAjax)
{
if (!modelStateValid)
{
return JsonProblemResult();
}
return Json();
}
if (!modelStateValid)
{
// this will throw and the exception page will be shown
throw new Exception();
}
return View();
}
Related
If I return StatusCode(403) or any other error code from an endpoint, any configuration of app.UseStatusCodePages<whatever> will be ignored.
I believe this is because the StatusCode(<whatever>) will automatically create a result object, and UseStatusCodePages only kicks in if there is an error status code and no content.
So how do I set a status code result in an IActionResult type endpoint and then return without setting any content so that UseStatusCodePages will handle the job of providing a suitable resonse?
As far as I know, the UseStatusCodePages will just be fired when the action result is the StatusCodeResult.
If you put some value inside the status codes, it will return the object result which will not trigger the UseStatusCodePages.
So I suggest you could directly use StatusCodeResult(403), then if you want to put some value to the StatusCodeResult, I suggest you could put it inside the httpcontext's item.
More details, you could refer to below codes:
public IActionResult OnGet()
{
HttpContext.Items.Add("test","1");
return StatusCode(403);
}
Program.cs:
app.UseStatusCodePages(async statusCodeContext =>
{
var status = statusCodeContext.HttpContext.Items["test"];
// using static System.Net.Mime.MediaTypeNames;
statusCodeContext.HttpContext.Response.ContentType = Text.Plain;
await statusCodeContext.HttpContext.Response.WriteAsync(
$"Status Code Page: {statusCodeContext.HttpContext.Response.StatusCode}");
});
Result:
The issue was that I have the ApiController attribute on the endpoint controller. One of the things this attribute does is to automatically create a ProblemDetails response body for any failed requests, and it is this that prevents UseStatusCodePages from having any effect.
The solution is to either remove the ApiController attribute if you do not require any of its features, or alternatively its behaviour of automatically creating ProblemDetails responses can be disabled using the following configuration in Program.cs (or Startup.cs in old style projects).
builder.Services.AddControllers().ConfigureApiBehaviorOptions(options =>
{
options.SuppressMapClientErrors = true;
});
In ASP.Net Core you have multiple ways to generate an URL for controller action, the newest being tag helpers.
Using tag-helpers for GET-requests asp-route is used to specify route parameters. It is from what I understand not supported to use complex objects in route request. And sometimes a page could have many different links pointing to itself, possible with minor addition to the URL for each link.
To me it seems wrong that any modification to controller action signature requires changing all tag-helpers using that action. I.e. if one adds string query to controller, one must add query to model and add asp-route-query="#Model.Query" 20 different places spread across cshtml-files. Using this approach is setting the code up for future bugs.
Is there a more elegant way of handling this? For example some way of having a Request object? (I.e. request object from controller can be put into Model and fed back into action URL.)
In my other answer I found a way to provide request object through Model.
From the SO article #tseng provided I found a smaller solution. This one does not use a request object in Model, but retains all route parameters unless explicitly overridden. It won't allow you to specify route through an request object, which is most often not what you want anyway. But it solved problem in OP.
<a asp-controller="Test" asp-action="HelloWorld" asp-all-route-data="#Context.GetQueryParameters()" asp-route-somestring="optional override">Link</a>
This requires an extension method to convert query parameters into a dictionary.
public static Dictionary GetQueryParameters(this HttpContext context)
{
return context.Request.Query.ToDictionary(d => d.Key, d => d.Value.ToString());
}
There's a rationale here that I don't think you're getting. GET requests are intentionally simplistic. They are supposed to describe a specific resource. They do no have bodies, because you're not supposed to be passing complex data objects in the first place. That's not how the HTTP protocol is designed.
Additionally, query string params should generally be optional. If some bit of data is required in order to identify the resource, it should be part of the main URI (i.e. the path). As such, neglecting to add something like a query param, should simply result in the full data set being returned instead of some subset defined by the query. Or in the case of something like a search page, it generally will result in a form being presented to the user to collect the query. In other words, you action should account for that param being missing and handle that situation accordingly.
Long and short, no, there is no way "elegant" way to handle this, I suppose, but the reason for that is that there doesn't need to be. If you're designing your routes and actions correctly, it's generally not an issue.
To solve this I'd like to have a request object used as route parameters for anchor TagHelper. This means that all route links are defined in only one location, not throughout solution. Changes made to request object model automatically propagates to URL for <a asp-action>-tags.
The benefit of this is reducing number of places in the code we need to change when changing method signature for a controller action. We localize change to model and action only.
I thought writing a tag-helper for a custom asp-object-route could help. I looked into chaining Taghelpers so mine could run before AnchorTagHelper, but that does not work. Creating instance and nesting them requires me to hardcode all properties of ASP.Net Cores AnchorTagHelper, which may require maintenance in the future. Also considered using a custom method with UrlHelper to build URL, but then TagHelper would not work.
The solution I landed on is to use asp-all-route-data as suggested by #kirk-larkin along with an extension method for serializing to Dictionary. Any asp-all-route-* will override values in asp-all-route-data.
<a asp-controller="Test" asp-action="HelloWorld" asp-all-route-data="#Model.RouteParameters.ToDictionary()" asp-route-somestring="optional override">Link</a>
ASP.Net Core can deserialize complex objects (including lists and child objects).
public IActionResult HelloWorld(HelloWorldRequest request) { }
In the request object (when used) would typically have only a few simple properties. But I thought it would be nice if it supported child objects as well. Serializing object into a Dictionary is usually done using reflection, which can be slow. I figured Newtonsoft.Json would be more optimized than writing simple reflection code myself, and found this implementation ready to go:
public static class ExtensionMethods
{
public static IDictionary ToDictionary(this object metaToken)
{
// From https://geeklearning.io/serialize-an-object-to-an-url-encoded-string-in-csharp/
if (metaToken == null)
{
return null;
}
JToken token = metaToken as JToken;
if (token == null)
{
return ToDictionary(JObject.FromObject(metaToken));
}
if (token.HasValues)
{
var contentData = new Dictionary();
foreach (var child in token.Children().ToList())
{
var childContent = child.ToDictionary();
if (childContent != null)
{
contentData = contentData.Concat(childContent)
.ToDictionary(k => k.Key, v => v.Value);
}
}
return contentData;
}
var jValue = token as JValue;
if (jValue?.Value == null)
{
return null;
}
var value = jValue?.Type == JTokenType.Date ?
jValue?.ToString("o", CultureInfo.InvariantCulture) :
jValue?.ToString(CultureInfo.InvariantCulture);
return new Dictionary { { token.Path, value } };
}
}
User requests page for Step1, fills out and submits form that contains selected person, so far so good. After validation of ModelState the next viewmodel is constructed properly using the selected person. I then attempt a redirect to action using the newVM but find on entry to Step2 that MVC wipes out the viewmodel attempted to be passed in. I suspect this is due to how MVC attempts to new up and instance based on query string results. I'll put a breakpoint in and check that, but am wondering how does one change a view from a post back with a new view model passed in?
public ActionResult Step1()
{
var vm = new VMStep1();
return View(vm);
}
[HttpPost]
public ActionResult Step1(VMStep1 vm)
{
if (ModelState.IsValid)
{
var newVM = new VMStep2(vm.SelectedPerson);
return RedirectToAction("Step2", newVM);
}
return View(vm);
}
public ActionResult Step2(VMStep2 vm)
{
return View(vm);
}
I can fix this by containing VMStep2 and a partial to Step2 in Step1 view, but that requires hide and seek logic when really I just want user to see Step2.
I don't see why you should want to call RedirectToAction! What it does it the following:
it tells your browser to redirect and like it or not your browser doesn't understand how to handle your object -- what it does understand is JSON. So if you really insist on using return RedirectToAction("Step2", newVM); you should consider a way to serialize your VMStep2 object to JSON and when the browser requests the Redirect, it will be properly passed and created in your action method public ActionResult Step2(VMStep2 vm)
HOWEVER I'd use a much simpler way ---
instead of
return RedirectToAction("Step2", newVM);
I would use
return View("Step2", newVM);
Thanks to everyone for the great input!
Here's what I did...
I created three views MainView, Step1View, Step2View (Step 1 and 2 were partial strong typed views)
I created a MainViewModel that contained VMStep1 and VMStep2
When controller served Step1 the MainViewModel only initialized VMStep1 and set state logic to tell MainView Step1 was to be shown.
When user posted back the MainView containing the MainViewModel, the MainViewModel knew what to do by the answers provided in VMStep1.
VMStep2 was initialized on the post back, and state was set to tell MainView to show Step2. VMStep1 was no longer relevant and was set to null.
User was now able to answer using VMStep2 and all was well.
The key to this working is that some flag tells the view which partial to show, the partial takes a model supporting it's strong type which is initialized at the right time. End result is fast rendering and good state machine progression.
Error like:The view 'LoginRegister' or its master was not found or no view engine supports the searched locations. The following locations were searched:
~/Views/MyAccount/LoginRegister.aspx
~/Views/MyAccount/LoginRegister.ascx
~/Views/Shared/LoginRegister.aspx
~/Views/Shared/LoginRegister.ascx
~/Views/MyAccount/LoginRegister.cshtml
~/Views/MyAccount/LoginRegister.vbhtml
~/Views/Shared/LoginRegister.cshtml
~/Views/Shared/LoginRegister.vbhtml
Actually my page view page is ~/Views/home/LoginRegister.cshtml so what i do
and my RouteConfig is
public class RouteConfig
{
public static void RegisterRoutes(RouteCollection routes)
{
routes.IgnoreRoute("{resource}.axd/{*pathInfo}");
routes.MapRoute(
name: "Default",
url: "{controller}/{action}/{id}",
defaults: new { controller = "MyAccount", action = "LoginRegister", id = UrlParameter.Optional }
);
}
}
Be careful if your model type is String because the second parameter of View(string, string) is masterName, not model. You may need to call the overload with object(model) as the second parameter:
Not correct :
protected ActionResult ShowMessageResult(string msg)
{
return View("Message",msg);
}
Correct :
protected ActionResult ShowMessageResult(string msg)
{
return View("Message",(object)msg);
}
OR (provided by bradlis7):
protected ActionResult ShowMessageResult(string msg)
{
return View("Message",model:msg);
}
Problem:
Your View cannot be found in default locations.
Explanation:
Views should be in the same folder named as the Controller or in the Shared folder.
Solution:
Either move your View to the MyAccount folder or create a HomeController.
Alternatives:
If you don't want to move your View or create a new Controller you can check at this link.
In Microsoft ASP.net MVC, the routing engine, which is used to parse incoming and outgoing URL Combinations, is designed with the idea of Convention over Configuration. What this means is that if you follow the Convention (rules) that the routing engine uses, you don't have to change the Configuration.
The routing engine for ASP.net MVC does not serve web pages (.cshtml). It provides a way for a URL to be handled by a Class in your code, which can render text/html to the output stream, or parse and serve the .cshtml files in a consistent manner using Convention.
The Convention which is used for routing is to match a Controller to a Class with a name similar to ControllerNameController i.e. controller="MyAccount" means find class named MyAccountController. Next comes the action, which is mapped to a function within the Controller Class, which usually returns an ActionResult. i.e. action="LoginRegister" will look for a function public ActionResult LoginRegister(){} in the controller's class. This function may return a View() which would be by Convention named LoginRegister.cshtml and would be stored in the /Views/MyAccount/ folder.
To summarize, you would have the following code:
/Controllers/MyAccountController.cs:
public class MyAccountController : Controller
{
public ActionResult LoginRegister()
{
return View();
}
}
/Views/MyAccount/LoginRegister.cshtml: Your view file.
In your LoginRegister action when returning the view, do below, i know this can be done in mvc 5, im not sure if in mvc 4 also.
public ActionResult Index()
{
return View("~/Views/home/LoginRegister.cshtml");
}
Check the build action of your view (.cshtml file) It should be set to content. In some cases, I have seen that the build action was set to None (by mistake) and this particular view was not deploy on the target machine even though you see that view present in visual studio project file under valid folder
This could be a permissions issue.
I had the same issue recently. As a test, I created a simple hello.html page. When I tried loading it, I got an error message regarding permissions. Once I fixed the permissions issue in the root web folder, both the html page and the MVC rendering issues were resolved.
Check whether the View (.ASPX File) that you have created is having the same name as mentioned in the Controller. For e.g:
public ActionResult GetView()
{
return View("MyView");
}
In this case, the aspx file should be having the name MyView.aspx instead of GetView.aspx
I got this error because I renamed my View (and POST action).
Finally I found that I forgot to rename BOTH GET and POST actions to new name.
Solution : Rename both GET and POST actions to match the View name.
If the problem happens intermittently in production, it could be due to an action method getting interrupted. For example, during a POST operation involving a large file upload, the user closes the browser window before the upload completes. In this case, the action method may throw a null reference exception resulting from a null model or view object. A solution would be to wrap the method body in a try/catch and return null. Like this:
[HttpPost]
public ActionResult Post(...)
{
try
{
...
}
catch (NullReferenceException ex) // could happen if POST is interrupted
{
// perhaps log a warning here
return null;
}
return View(model);
}
I had this same issue.
I had copied a view "Movie" and renamed it "Customer" accordingly.
I also did the same with the models and the controllers.
The resolution was this...I rename the Customer View to Customer1 and
just created a new view and called it Customer....I then just copied
the Customer1 code into Customer.
This worked.
I would love to know the real cause of the problem.
UPDATE
Just for grins....I went back and replicated all the renaming scenario again...and did not get any errors.
I came across this error due to the improper closing of the statement,
#using (Html.BeginForm("DeleteSelected", "Employee", FormMethod.Post))
{
} //This curly bracket needed to be closed at the end.
In Index.cshtml view file.I didn't close the statement at the end of the program. instead, I ended up closing improperly and ran into this error.
I was sure there isn't a need of checking Controller ActionMethod code because I have returned the Controller method properly to the View. So It has to be the view that's not responding and met with similar Error.
If you've checked all the things from the above answers (which are common mistakes) and you're sure that your view is at the location in the exceptions, then you may need to restart Visual Studio.
:(
In my case, I needed to use RedirectToAction to solve the problem.
[HttpGet]
[ControleDeAcessoAuthorize("Report/ExportToPDF")]
public ActionResult ExportToPDF(int id, string month, string output)
{
try
{
// Validate
if (output != "PDF")
{
throw new Exception("Invalid output.");
}
else
{
...// code to generate report in PDF format
}
}
catch (Exception ex)
{
return RedirectToAction("Error");
}
}
[ControleDeAcessoAuthorize("Report/Error")]
public ActionResult Error()
{
return View();
}
I ran into this a while ago and it drove me crazy because it turned out to be simple. So within my View I was using a grid control that obtained data for the grid via an http request. Once the middle tier completed my request and returned the dataset, I received the same error. Turns out my return statement was 'return View(dataset);' instead of 'return Json(dataset);
I couldn't find any solution to this problem, until I found out the files didn't exist!
This took me a long time to figure out, because the Solution Explorer shows the files!
But when I click on Index.cshtml I get this error:
So that was the reason for this error to show. I hope this answer helps somebody.
I am currently working on an Umbraco MVC 4 project version 6.0.5. The project currently uses Vega.USiteBuilder to build the appropriate document types in the backoffice based on strongly typed classes with mapping attributes. Consequently, all my razor files inherit from UmbracoTemplatePageBase
I am coming across a road block trying to invoke a HTTP GET from a razor file. For example a search form with multiple fields to submit to a controller action method, using a SurfaceController using Html.BeginUmbracoForm.
My Html.BeginUmbracoForm looks like this
#using (Html.BeginUmbracoForm("FindTyres", "TyreSearch"))
{
// Couple of filter fields
}
I basically have a scenario where I will like to retrieve some records from an external database outside of Umbraco (external to Umbraco Database) and return the results in a custom view model back to my Umbraco front end view. Once my controller and action method is setup to inherit from SurfaceController and thereafter compiling it and submitting the search, I get a 404 resource cannot be found where the requested url specified: /umbraco.RenderMVC.
Here is my code snippet:
public ActionResult FindTyres(string maker, string years, string models, string vehicles)
{
var tyreBdl = new Wheels.BDL.TyreBDL();
List<Tyre> tyres = tyreBdl.GetAllTyres();
tyres = tyres.Where(t => string.Equals(t.Maker, maker, StringComparison.OrdinalIgnoreCase)
&& string.Equals(t.Year, years, StringComparison.OrdinalIgnoreCase)
&& string.Equals(t.Model, models, StringComparison.OrdinalIgnoreCase)
&& string.Equals(t.Version, vehicles, StringComparison.OrdinalIgnoreCase)).ToList();
var tyreSearchViewModel = new TyreSearchViewModel
{
Tyres = tyres
};
ViewBag.TyreSearchViewModel = tyreSearchViewModel;
return CurrentUmbracoPage();
}
I then resort to using standard MVC, Html.BeginForm (the only difference). Repeating the steps above and submitting the search, I get the following YSOD error.
Can only use UmbracoPageResult in the context of an Http POST when
using a SurfaceController form
Below is a snippet of the HTML BeginForm
#using (Html.BeginForm("FindTyres", "TyreSearch"))
{
// Couple of filter fields
}
I feel like I am fighting the Umbraco routes to get my controller to return a custom model back to the razor file. I have googled alot trying to figure out how to do a basic search to return a custom model back to my Umbraco front end view till the extent that I tried to create a custom route but that too did not work for me.
Does my controller need to inherit from a special umbraco controller class to return the custom model back? I will basically like to invoke a HTTP GET request (which is a must) so that my criteria search fields are reflected properly in the query strings of the url. For example upon hitting the search button, I must see the example url in my address browser bar
http://[domainname]/selecttyres.aspx/TyresSearch/FindTyresMake=ASIA&Years=1994&Models=ROCSTA&Vehicles=261
Therefore, I cannot use Surface Controller as that will operate in the context of a HTTP Post.
Are there good resource materials that I can read up more on umbraco controllers, routes and pipeline.
I hope this scenario makes sense to you. If you have any questions, please let me know. I will need to understand this concept to continue on from here with my project and I do have a deadline.
There are a lot of questions about this and the best place to look for an authoritative approach is the Umbraco MVC documentation.
However, yes you will find, if you use Html.BeginUmbracoForm(...) you will be forced into a HttpPost action. With this kind of functionality (a search form), I usually build the form manually with a GET method and have it submit a querystring to a specific node URL.
<form action="#Model.Content.Url"> ... </form>
On that page I include an #Html.Action("SearchResults", "TyresSearch") which itself has a model that maps to the keys in the querystring:
[ChildAction]
public ActionResult(TyreSearchModel model){
// Find results
TyreSearchResultModel results = new Wheels.BDL.TyreBDL().GetAllTyres();
// Filter results based on submitted model
...
// Return results
return results;
}
The results view just need to have a model of TyreSearchResultModel (or whatever you choose).
This approach bypasses the need for Umbraco's Controller implementation and very straightforward.
I have managed to find my solution through route hijacking which enabled me to return a custom view model back to my view and work with HTTP GET. It worked well for me.
Digby, your solution looks plausible but I have not attempted at it. If I do have a widget sitting on my page, I will definitely attempt to use your approach.
Here are the details. I basically override the Umbraco default MVC routing by creating a controller that derived from RenderMvcController. In a nutshell, you implement route hijacking by implementing a controller that derives from RenderMvcController and renaming your controllername after your given documenttype name. Recommend the read right out of the Umbraco reference (http://our.umbraco.org/documentation/Reference/Mvc/custom-controllers) This is also a great article (http://www.ben-morris.com/using-umbraco-6-to-create-an-asp-net-mvc-4-web-applicatio)
Here is my snippet of my code:
public class ProductTyreSelectorController : Umbraco.Web.Mvc.RenderMvcController
{
public override ActionResult Index(RenderModel model)
{
var productTyreSelectorViewModel = new ProductTyreSelectorViewModel(model);
var maker = Request.QueryString["Make"];
var years = Request.QueryString["Years"];
var models = Request.QueryString["Models"];
var autoIdStr = Request.QueryString["Vehicles"];
var width = Request.QueryString["Widths"];
var aspectRatio = Request.QueryString["AspectRatio"];
var rims = Request.QueryString["Rims"];
var tyrePlusBdl = new TPWheelBDL.TyrePlusBDL();
List<Tyre> tyres = tyrePlusBdl.GetAllTyres();
if (Request.QueryString.Count == 0)
{
return CurrentTemplate(productTyreSelectorViewModel);
}
if (!string.IsNullOrEmpty(maker) && !string.IsNullOrEmpty(years) && !string.IsNullOrEmpty(models) &&
!string.IsNullOrEmpty(autoIdStr))
{
int autoId;
int.TryParse(autoIdStr, out autoId);
tyres = tyres.Where(t => string.Equals(t.Maker, maker, StringComparison.OrdinalIgnoreCase) &&
string.Equals(t.Year, years, StringComparison.OrdinalIgnoreCase) &&
string.Equals(t.Model, models, StringComparison.OrdinalIgnoreCase) &&
t.AutoID == autoId)
.ToList();
productTyreSelectorViewModel.Tyres = tyres;
}
else if (!string.IsNullOrEmpty(width) && !string.IsNullOrEmpty(aspectRatio) && !string.IsNullOrEmpty(rims))
{
tyres = tyres.Where(t => string.Equals(t.Aspect, aspectRatio, StringComparison.OrdinalIgnoreCase) &&
string.Equals(t.Rim, rims, StringComparison.OrdinalIgnoreCase)).ToList();
productTyreSelectorViewModel.Tyres = tyres;
}
var template = ControllerContext.RouteData.Values["action"].ToString();
//return an empty content result if the template doesn't physically
//exist on the file system
if (!EnsurePhsyicalViewExists(template))
{
return Content("Could not find physical view template.");
}
return CurrentTemplate(productTyreSelectorViewModel);
}
}
Note my ProductTyreSelectorViewModel must inherit from RenderModel for this to work and my document type is called ProductTyreSelector. This way when my model is returned with the action result CurrentTemplate, the Umbraco context of the page is retained and my page is rendered appropriately again. This way, all my query strings will show all my search/filter fields which is what I want.
Here is my snippet of the ProductTyreSelectorViewModel class:
public class ProductTyreSelectorViewModel : RenderModel
{
public ProductTyreSelectorViewModel(RenderModel model)
: base(model.Content, model.CurrentCulture)
{
Tyres = new List<Tyre>();
}
public ProductTyreSelectorViewModel(IPublishedContent content, CultureInfo culture)
: base(content, culture)
{
}
public ProductTyreSelectorViewModel(IPublishedContent content)
: base(content)
{
}
public IList<Tyre> Tyres { get; set; }
}
This approach will work well perhaps with one to two HTTP GET forms on a given page. If there are multiple forms within in a page, then a good solution will may be to use ChildAction approach. Something I will experiment with further.
Hope this helps!