I am looking at an example of how to use MVC Contrib Grid :
https://web.archive.org/web/20211020001718/https://www.4guysfromrolla.com/articles/031611-1.aspx
I am already using a #model on the page for something else, and I want to be able to create up to 6 grids on my page. Is it possible to pass a method that returns the grid contents to the Grid object?
You can pass any number of collections from the controller action to the view through ViewData or ViewBag.
public ActionResult Index()
{
ViewData["foos"] = GetFoos();
ViewData["bars"] = GetBars();
ViewData["bazz"] = GetBazz();
return View(/*model*/);
}
And in the view
#Html.Grid((IEnumerable<Foo>)ViewData["foos"]).Columns(...)
#Html.Grid((IEnumerable<Bar>)ViewData["bars"]).Columns(...)
#Html.Grid((IEnumerable<Bazz>)ViewData["bazz"]).Columns(...)
Related
MVC/ASP.NET/C#/html/javascript newbie question:
I'm trying to move some legacy software into an MVC solution. I have an MVC controller ViewResult method that makes an API call to the legacy system and returns a string which is a fully formed HTML page (including the HTML start and end tags). Some time in the future, I'll rewrite the logic as an MVC view, but for right now I need to just display that page (preferably in a new tab).
I've tried this in the controller:
return View((object)calendar);
(where "calendar" is the string containing the HTML document)
In my view I have
#model string
#{ Layout = null; }
#Model
But that didn't work.
Any ideas?
Model binding is binding the object of your model class.
For example, ([Solution].[Models].[Model class]),
#model PassDatainMVC.Models.Record
To pass the data from controller to view,
Approach 1: ViewBag
Controller:
string data = "testing";
ViewBag.see = data;
return View();
View:
#using PassDatainMVC.Models
#ViewBag.see
Or:
Approach 2: Model binding
Controller (Class):
public string recordProperty;
View:
#model PassDatainMVC.Models.Record
#Model.recordProperty
While you have to set the property under the model class in the data field for the second approach.
Ref. https://www.c-sharpcorner.com/article/asp-net-mvc-passing-data-from-controller-to-view/
If you want to just one data you can use a ViewBag. This is simple.
Also you want to send with model. You should use this code.
Class
public class Calendar
{
public string CalendarName { get; set; }
}
Controller
Calendar newModel = new Calendar();
newModel.CalendarName = "test name...";
return View(newModel);
View
#model ModelNamespace.Calendar
<h1> #Model.CalendarName </h1>
Thanks Reha! But unfortunately neither of your suggestions did the trick.
For your first suggestion I used ViewBag. In the controller I replaced
return View((object)calendar);
to
ViewBag.calendar = calendar;
return View();
And replaced the view with just
#{ Layout = null; }
#ViewBag.calendar
The result was that the user is left looking at the actual HTML code instead of what the HTML code is supposed to render.
For your 2nd suggestion, I did exactly as you suggested (except I changed
Model.CalendarName = "test name...";
to
Model.CalendarName = calendar;
The result is the same, the user is left looking at the HTML code.
I'm currently generating breadcrumbs on an object's Details page by calling a GetBreadcrumbs() method in the object's controller - in this method, the object's parent/grandparent are used to generate an unordered list. What's the best way to pull the HTML out of the controller to follow the Separation of Concerns paradigm? Should a partial view be used here?
Typical example of partial view is Breadcrum itself. For example, in your controller you can have
//
//GET: News/Article/x
public ActionResult Article(int id)
{
//get parentid of article
ViewBag.id = id;
ViewBag.parentid;
return View();
}
So, your partial view will be as below:
#{
ViewBag.Title = "Article";
}
<h2>Viewing Article #ViewBag.parentid >> #ViewBag.id</h2>
You could use partial views or display templates. Your controller should only build the model that will be passed to the view and then inside the view you could use a display template that will build the desired output based on the model.
I currently have a _layout.cshtml used by every page of my website.
I need to put a form on each page displayed as a popin.
So, i created a new PartialView (the content of my form) with its corresponding ViewModel and called it in _layout.cshtml.
However, i have a model conflict between ViewModels of pages using the layout and the ViewModel used by the new form (since we can't have directly two models for the same view).
The model item passed into the dictionary is of type 'XXX', but this
dictionary requires a model item of type 'YYY'.
How can we include a form in _layout without this conflict ?
The following has worked for me with a sidebar on every page.
Create a controller for your partial view
In that controller, create a method for the view you want to return, and be sure to use the [ChildActionOnly] filter
public class PartialController : Controller
{
[ChildActionOnly]
public PartialViewResult Alerts()
{
return PartialView("Alerts", messages);
}
}
In your _layout view, you'll have the following:
#Html.Action("Alerts", "Partial")
(instead of #Html.RenderPartial or #Html.Partial)
It sounds like you already have what you need for the view.
I have not used this with a form, but it should work similarly. Hope this helps.
Using the EditorFor templates is a really nice feature of ASP.Net MVC 3, but is it possible to get EditorFor to render an unpopulated template to allow for creation of records?
Or is there some other way to do this?
The ways in which I am trying to do this is as follows:
#Html.EditorFor(model => model)
#Html.EditorFor(x => new List<Business.ViewModel.Affiliate.Contact>())
#Html.EditorFor(new List<Business.ViewModel.Affiliate.Contact>())
#Html.EditorFor(new Business.ViewModel.Affiliate.Contact())
The first one obviously works, however the subsequent ones (which demonstrate what I am trying to do) all fail with the following error:
Templates can be used only with field access, property access, single-dimension array index, or single-parameter custom indexer expressions.
The model in question is:
IEnumerable<Business.ViewModel.Affiliate.Contact>
It's the responsibility of the controller to prepare the view model that will be passed to the view. So if you need for example to initialize your view view model with 5 empty contact rows you could do this simply in your controller:
public ActionResult Index()
{
var model = new MyViewModel
{
// Add 5 empty contacts
Contacts = Enumerable.Range(1, 5).Select(x => new Contact()).ToList()
};
return View(model);
}
and in your view use the EditorFor helper as usual:
#model MyViewModel
...
#Html.EditorFor(x => x.Contacts)
This will render the corresponding editor template for each of the 5 elements we have added to the Contacts collection.
If your question doesn't involve AJAX, then I would design the ViewModel as following:
class MyList
{
public List<MyRow> Rows {get;set;}
public MyRow NewRow {get;set;}
}
Then you can easily add a blank editor bound to NewRow property. And in the controller you add the NewRow to Rows on subsequent calls.
In my asp.net mvc3 application i have created two partial views for two different action that is,
partialviewresult setcomment and
partialviewresult getcomment
i have created partial view using create a strongly type view and different scaffold template
for _setcomment i am using create template and for _getcomment i am using List template.
Now i want to call both _setcomment and _getcomment partial view in one view.
in my view file .cshtml
_setcomment -
#model <NAMESPACE>.<MODELNAME>
<some code>
_getcomment -
#model IEnumerable<<NAMESPACE>.<MODELNAME>>
<some code>
how can i call diiferent partial view in one view?
any suggestions?
There are different ways to do it.
If you already have the model class data in the Main view you can use like
In the main view call
#Html.Partial("PartialViewName1",model1)
#Html.Partial("PartialViewName1",model2)
If you do not have the model class data in the mail view then you can call the action on the controller and from there return the partial view.
#Html.Action("Controller","Action1")
#Html.Action("Controller","Action2")
In the Controller class
PartialResult Action1()
{
model = new ModelClass();
return PartialView(model);
}
Hope this helps.
The answer to your question is to use the following within a single view:
#{ Html.RenderAction("ActionName", "ControlerName"); }
#{ Html.RenderAction("ActionName2", "ControlerName2"); }
This would do what you are trying to achieve, however, I think there is a problem with design. What are you trying to achieve?