I have a person class that has an index file that lists each person.
An Html.ActionLink points to another controller for notes on the person.
One person, many notes, so I want to pass the PersonID as a param and insert it into the new note form. The parameter is not the NoteID, ie the key
#Html.ActionLink("Note", "Create", "Note", new { id = item.PersonID }, null)
The PersonID is in the url that passes to the note form.
How do I get the PersonID into the Note form?
Many thanks,
Harry
In your Create action method in the Note controller, you will receive the ID as a paramter to the action method if you include a parameter for it in the function:
public ActionResult Create(string id)
{
}
From there it is up to you to pass the id into the view you are using to construct the Create Note page. If you have a Note model object and one of the properties is the PersonID, you might want to new up a new Note object, set the PersonID to the value passed in the parameter, and then pass the new Note object to the view page as the model:
return View(newNote);
Hope that is what you are looking for.
You can knockout it using the same name as u declared ID
Or.. just handle it on another Controller using
the same name. ID:
[HttpGet]
public ViewResult Create(int id)
{
return View(id); <--
}
Then with razor-engine it comes with #Model
Related
I can't figure out why I need to create an empty method signature to allow a Get rest call with 3 null parameters to work. I have the following Code:
public class SessionPresenterController : ApiController
{
public HttpResponseMessage Get()
{
return Get(null, null, null);
}
public HttpResponseMessage Get(int? codeCampYearId, int? sessionId, int? attendeesId)
{
and in my WebApiConfig I have
config.Routes.MapHttpRoute
("API Default Rest", "rest/{controller}/{id}",
new { id = RouteParameter.Optional });
config.Routes.MapHttpRoute
("API Default RPC", "rpc/{controller}/{action}/{id}",
new { id = RouteParameter.Optional });
If I call /rest/SessionPresenter without any parameters and I Don't have the Get() defined, the Get with the three null parameters is not found.
Why not?
Try specifying defaults for your parameters:
public HttpResponseMessage Get(
int? codeCampYearId = null,
int? sessionId = null,
int? attendeesId = null)
{
//...
}
Jacobs' snippet will solve your issue. Let's answert the question why not? why get is not found..
There is a complete documentation of the Routing and Action Selection. Let's use some extratc and reveal what happens:
Action Selection
Create a list of all actions on the controller that match the HTTP request method.
If the route dictionary has an "action" entry, remove actions whose name does not match this value.
Try to match action parameters to the URI, as follows:
For each action, get a list of the parameters that are a simple type, where the binding gets the parameter from the URI. Exclude optional parameters.
From this list, try to find a match for each parameter name, either in the route dictionary or in the URI query string. Matches are case insensitive and do not depend on the parameter order.
Select an action where every parameter in the list has a match in the URI.
If more that one action meets these criteria, pick the one with the most parameter matches.
Other words, if there will be only Get with three parameters (omited Get()), to decided which action should be selected:
Selected was Get(int? codeCampYearId, int? sessionId, int? attendeesId)
still the Get(int? codeCampYearId, int? sessionId, int? attendeesId) is selected
URL is /rest/SessionPresenter
no optional parameters excluded. All have to be found
URL does not have a match for all three parameters
So to solve it, we have to either pass all params (empty, null):
/rest/SessionPresenter?codeCampYearId&attendeesId&sessionid
Or change the signature to have parameters optinal (Jacobs' answer), and skipped during the action selection
I known if I have something like /controller/action/{id} I can access id as a function parameter. But how would I access it via the view without using the viewbag?
You can pass that function parameter to the View as part of a model. The model can be staticly or dynamically typed. The example code below demonstrates how to pass the value as a property on a dynamic model.
public ActionResult Edit(string id)
{
dynamic model = new System.Dynamic.ExpandoObject();
model.Id = id;
return View(model);
}
You would access this value in the view as follows:
#Model.Id
You can parse the URL to get the ID when in the view:
var id = Request.Url.LocalPath.SubString(LastIndexOf("/",Request.Url.LocalPath)+1);
It will be put directly into the ViewBag.id or you can access it via ViewData["id"]. hey seem like the same object
How do I pass a whole model via html.actionlink or using any other method except form submission? Is there any way or tips for it?
Though it's not advisable in complex cases, you can still do that!
public class QueryViewModel
{
public string Search { get; set; }
public string Category { get; set; }
public int Page { get; set; }
}
// just for testing
#{
var queryViewModel = new QueryViewModel
{
Search = "routing",
Category = "mvc",
Page = 23
};
}
#Html.ActionLink("Looking for something", "SearchAction", "SearchController"
queryViewModel, null);
This will generate an action link with href like this,
/SearchController/SearchAction?Search=routing&Category=mvc&Page=23
Here will be your action,
public ViewResult SearchAction(QueryViewModel query)
{
...
}
No, you cannot pass entire complex objects with links or forms. You have a couple of possible approaches that you could take:
Include each individual property of the object as query string parameters (or input fields if you are using a form) so that the default model binder is able to reconstruct the object back in the controller action
Pass only an id as query string parameter (or input field if you are using a form) and have the controller action use this id to retrieve the actual object from some data store
Use session
You could use javascript to detect a click on the link, serialize the form (or whatever data you want to pass) and append it to your request parameters. This should achieve what you're looking to achieve...
I have this post method:
[HttpPost]
[ValidateAntiForgeryToken]
public ActionResult Invitations(SuperInvitationsEditModel model)
{
...
var newmodel = new SuperInvitationsEditModel();
if (hasErrors)
{
SuperInvitationsErrorModel newErrorModel = new SuperInvitationsErrorModel();
newErrorModel.Errors = model.Errors;
return View(newErrorModel);
}
return View(newmodel);
}
When this code in the if(hasErrors) executes I get this error.
The model item passed into the dictionary is of type 'MyProject.Models.SuperInvitationsErrorModel', but this dictionary requires a model item of type 'MyProject.Models.SuperInvitationsEditModel'.
I thought I can do this since the return value of the method is a generic ActionResult. Can anyone tell me why is this not working?
because your current view is strongly typed. change the code as
return View("yourviewname",newErrorModel);
It has nothing to do with casting ViewResult to ActionResult. The problem is, that you have strongly typed view that expects the model of type SuperInvitationsEditModel (see #model on the top of Invitations.cshtml), but you are passing the model of type SuperInvitationsErrorModel to it.
You should merge the two view model classes (SuperInvitationsEditModel and SuperInvitationsErrorModel) into one, or create a standalone view for each of them.
I have an action that creates a List and returns it to my view..
public ActionResult GetCustomers()
{
return PartialView("~/Views/Shared/DisplayTemplates/Customers.cshtml", UserQueries.GetCustomers(SiteInfo.Current.Id));
}
And in the "~/Views/Shared/DisplayTemplates/Customers.cshtml" view I have the following:
#model IEnumerable<FishEye.Models.CustomerModel>
#Html.DisplayForModel("Customer")
Then I have in the "~/Views/Shared/DisplayTemplates/Customer.cshtml" view:
#model FishEye.Models.CustomerModel
#Model.Profile.FirstName
I am getting the error:
The model item passed into the dictionary is of type System.Collections.Generic.List`1[Models.CustomerModel]', but this dictionary requires a model item of type 'Models.CustomerModel'.
Shouldn't it display the Customer.cshtml for every item in the collection in the Customers.cshtml?
Help!
I am not sure why you are calling a partial view like this. If it is a Customer Specific view, why not put it under Views/Customer folder ? Remember ASP.NET MVC is more of Conventions. so i would always stick with the conventions (unless abosultely necessary to configure myself) to keep it simple.
To handle this situation, i would do it in this way,
a Customer and CustomerList model/Videmodel
public class CustomerList
{
public List<Customer> Customers { get; set; }
//Other Properties as you wish also
}
public class Customer
{
public string Name { get; set; }
}
And in the action method, i would return an object of CustomerList class
CustomerList customerList = new CustomerList();
customerList.Customers = new List<Customer>();
customerList.Customers.Add(new Customer { Name = "Malibu" });
// you may replace the above manual adding with a db call.
return View("CustomerList", customerList);
Now there should be a view called CustomerList.cshtml under Views/YourControllerName/ folder. That view should look like this
#model CustomerList
<p>List of Customers</p>
#Html.DisplayFor(x=>x.Customers)
Have a view called Customer.cshtml under Views/Shared/DisplayTemplates with this content
#model Customer
<h2>#Model.Name</h2>
This will give you the desired output.
Your view is expecting a single model:
#model FishEye.Models.CustomerModel // <--- Just one of me
You're passing it an anonymous List:
... , UserQueries.GetCustomers(SiteInfo.Current.Id) // <--- Many of me
You should change your view to accept the List or determine which item in the list is supposed to be used before passing it into the View. Keep in mind, a list with 1 item is still a list and the View is not allowed to guess.