how to catch id in aspx page in MVC - c#

I'm working on a MVC production project.
In my Production details view I have some buttons to get some more data from the database, but for this I need the id of the Product. I can see it exist but can I catch it?
Here's my controller that return data:
public ActionResult Details(long AProductionOrderId)
{
ProductionOrderList item = new ProductionOrderList();
item = ProductionOrderReg.GetProductionOrders(conn, AProductionOrderId);
ViewData["item"] = item;
return View();
}
Here's my details page when it load, I can see the id, but how to catch and use it in the buttons in the left to bring more date ?

You could use a hidden input on your view page to submit the ID.
your View:
<form method="post">
<button type="submit">Button Text</button>
<input type="hidden" name="AProductionOrderId" value="#ViewData['item']">
</form>

i wrote this im my controller
ViewData["id"] = AProductionOrderId;
and catched it in my view
long id = Convert.ToInt64( ViewData["id"]);

If you controller is:
public ActionResult Details(long AProductionOrderId)
{
var item = ProductionOrderReg.GetProductionOrders(conn, AProductionOrderId);
ViewBag.ProductionOrderId = AProductionOrderId;
return View(item);
}
then your AProductionOrderId will be in the ViewBag although I don't see the reason why you need it since whatever the type of item is (single object instance or list of objects) it contains your ID as a property because you're fetching the item by this ID. Anyway in your model you then need to declare your model like this:
#model YourModelNamespace.ProductionOrderList
and now you can access any property of your model in your view. But if you really want you can access it via ViewBag like this:
#{
long AProductionOrderId = Viewbag.AProductionOrderId;
}

Related

How can i send back a model containing two lists from the view to my controller

I have a cshtml view, for which I am sending a viewmodel. That viewmodel consist of two list of albums (music albums). I then ask the user to check up to 3 of these albums (a checkbox next to the title) to vote for his favorite music. I use javascript to ensure he doesn't check anymore than 3 (The security is a detail right now, I'm more concerned about getting it to work, but I'm open to suggestion if people have a better solution).
Since all albums are displayed in a table, I would love to send back to the controller through the submit button, the same model after updating it.
Basically, one of the list contain the current vote the user has made before loading the page (can be empty), and the second one should be empty until sent back to the controller containing the list of votes that are currently selected. I then use these two lists to compare them and update the database, removing the votes he removed, and adding the vote he added.
But I am unable to create a proper form to return these informations as I am not used to forms.
I tried to put the whole list in a form, but it didn't work. My reserach when I look for "sending model back to controller" usually do just that and get it to work.
View model
public class CategoryVotesUserViewModels
{
public CategoryVoteViewModels categoryVoteViewModels;
public List<int> listVotesEntry = new List<int>();
public List<int> listVotesOutput = new List<int>();
}
Relevant CSHTML and javascript
#section Header{
<script>
var MAX_VOTES = 3;
function checkNumberVotes($this) {
console.log($("input[name='listVoteOutput']:checked"));
if ($("input[name='listVoteOutput']:checked").length > MAX_VOTES) {
$this.checked = false;
}
}
</script>
}
#using (Html.BeginForm("VoteInCategory", "Votes", new { SearchModel = Model }, FormMethod.Post))
{
<ul>
#foreach (var av in Model.categoryVoteViewModels.listVotes)
{
<li>
#av.album.Title | #av.votes |
<input type="checkbox"
name="listVoteOutput"
value=#av.album.ID
onclick="checkNumberVotes(this)"
#if (Model.listVotesEntry.Contains(av.album.ID))
{ <text> checked </text> } />
</li>
}
</ul>
<div class="form-group text-center">
<input type="submit" class="btn btn-primary" value="Submit" />
</div>
}
Controller
[HttpPost]
public ActionResult VoteInCategory(CategoryVotesUserViewModels categoryVotesUserViewModels)
{
if (ModelState.IsValid)
{
List<int> toAdd = categoryVotesUserViewModels.listVotesOutput.Except(categoryVotesUserViewModels.listVotesEntry).ToList();
List<int> toRemove = categoryVotesUserViewModels.listVotesEntry.Except(categoryVotesUserViewModels.listVotesOutput).ToList();
VoteService.updateVoteUserCategory(User.Identity.GetUserId(), toRemove, toAdd, categoryVotesUserViewModels.categoryVoteViewModels.categoryID);
//TODO Redirect to success
return RedirectToAction("Index", "Home");
}
return View(categoryVotesUserViewModels);
}
If the user already had voted, all album whose ID is in "ListVotesEntry" should begin checked. If the user hasn't voted, or voted for nothing previously, "ListVotesEntry" should be empty.
When the User press the submit button, if an album is checked, the album's id should be added to the "ListVotesOutput" list. Also, both "ListVotesEntry" and "ListVotesOutput" should be sent back to the controller. The list with the names of the albums and their titles/ID is no longer necessary for the rest of the treatment
Found the solution. The problem was that my model needed to use "{get; set;}" on its attributes, otherwise the binding doesn't work, which mean that it send back an empty model.

Return json to partial view in core 2

I would like to create a view that would contain a different view. I've never used json before. How i can do this and How can I format the json data in the view?
My first function "Details" is to retrieve a object from the database and return view "Details.cshtml". In this view I want generates a partial view ("Stats.cshtml"). And now I want to generate a partial view with the data downloaded in the json format inside the Stats function.
Controller
public IActionResult Details(int? id = 1)
{
var person = _context.Persons.Find(id);
return View(champion);
}
public IActionResult Stats()
{
var json = new WebClient().DownloadString("url");
return Json(s);
}
View - Details.cshtml
#model Person
<div class=row">
<div class="col-sm-5"> #Model.Name </div>
<div class="col-sm-5"> #Html.Partial("Stats") </div>
</div>
View - Stats.cshtml
<h2>Stats</h2>
<div> here I want to put in a json field </div>
When I run "Stats" function from the address localhost/Home/Stats I get the result in json, but when I run "Details" function I get view "Details" and "Stats" without the json value.
to render a partial, you have many options, by your code,
the simplest one is: move your Stats code to Details action
public ActionResult Details()
{
...//prepare your person viewModel
var result = new WebClient().DownloadString("url");
var stats = JsonConvert.DeserializeObject<YourViewModel>(result);
//you have 2 options to return data
yourPersonModel.Stats=stats ; //<== you have to change PersonViewModel
//or ViewBag.Stats=stats;
return View(yourPersonModel);
}
then in Details.cshtml:
#Html.Partial("Stats", ViewBag.Stats or Model.Stats)//by your choice before.
Since Html.Action is removed, but ViewComponent comes in Core, you cannot directly call it right now, but this link will tell you how to make it back: #Html.Action in Asp.Net Core
public ActionResult Stats()
{
var result = new WebClient().DownloadString("url");
var yourViewModel = JsonConvert.DeserializeObject<YourViewModel>(result);
return PartialView(yourViewModel);
}
add the following code in your View - Stats.cshtml:
#model YourViewModel
then in Details.cshtml:
#Html.Action("Stats")
Be aware that Html.Action cannot call async actions, be careful to use it.
the next solution is to use new feature ViewComponent, here is details:
https://learn.microsoft.com/en-us/aspnet/core/mvc/views/view-components?view=aspnetcore-2.1
the last one will not be what you expected: use AJAX to load this partial page on page Details is loaded to client

Get RadioButton Value KnockoutJs

I have dynamically created radio buttons in my view and i am trying to pass the checked value to my controller. When my controller is hit the Agentcode string is Empty and i cant figure out how to grab the value so i can send it to my controller.
EDIT my radio buttons are generated in a foreach which i think may be causing the difficulty in reaching the value.
This is what the html rendered by the MVC control looks like
<div>
<input name="XXXXX" type="radio" value="{ data_bind = checkedValue: $data,checked: $root.AgentCode }">
</div>
foreach (var code in Model.ActiveAgentCodes)
{
<div>
#Html.RadioButton(code.AgentCode, new { data_bind="checkedValue: $data,checked: $root.AgentCode"})
#Html.RadioButton(code.AgentCode, new {data_bind="checkedValue: $data,checked: $root.AgentCode"}) }
my knockout ViewModel looks like this.
function ViewModel(){
var self = this;
self.AgentCode = ko.observable();
};
var viewModel = new ViewModel();
ko.applyBindings(viewModel);
and the post method in my controller looks like this
[HttpPost]
public ActionResult GetAgentCodeForHomeController(string AgentCode)
{
return RedirectToAction("Home", "Home");
}
In my view i am posting using like so
#using (Html.BeginForm("GetAgentCodeForHomeController", "ChangeAccount"))
{
#Html.RadioButton(code.AgentCode, new {data_bind="checkedValue: $data,checked: $root.AgentCode"})
#Html.HiddenFor(model => model.AgentCode, new { data_bind = "text:AgentCode" })
}
<button type="submit">OK</button>
I needed to send data back to my controller and the best way that i found was to add properties to my viewmodel.
public string AgentCode {get; set;}
Then give the radio buttons an Id so that when the controller is hit MVC maps the properties correctly. I decided to not go with knockout and instead post the form back to the controller. It ended up looking like this on my buttons.
#Html.RadioButton("AgentCode", code.AgentCode, new {id = code.AgentCode})

Best way to handle add/view/delete on one page

What I want to do
I am very new to MVC.
I'm trying to create a page that allows users to perform the following actions on the same page:
View the list (table)
Add a new item (Filling the form and clicking the Add button should update the table)
Delete an item from the list (Clicking the Delete button in a row should update the table)
A simple example looks like this but I actually have two lists on one page (Fees and Costs):
Question
What would be the best way to achieve this?
Should I go with Dylan Beattie's method posted here which would look something like this?
public ActionResult MyAction(string submitButton, MyViewModel form)
{
switch (submitButton)
{
case "AddFee":
return (AddFee(form));
case "AddCost":
return (AddCost(form));
case "RemoveFee":
return (RemoveFee(form));
case "RemoveCost":
return (RemoveCost(form));
}
}
public ActionResult AddFee(MyViewModel form)
{
Fee newFee = ....; // Get entered data from `form`
_repository.InsertFee(newFee);
return View("Create"); //Back to the original page
}
Or is there any other recommended methods to handle this such as using JavaScript?
You could create the table as a partial view and re render this via ajax.
Wrap the partial view in a div and Wrap the form in #using (Ajax.BeginForm(.... and target the wrapper div. Your controller action that is targeted by the ajax request will need to return a partial view.
Here is a simple example
public class HomeController : Controller
{
public ActionResult Index()
{
MYvm vm = new MYvm() { id = 1, name = "This is my View Model" };
return View(vm);
}
public ActionResult DA(MYvm vm)
{
vm.name = "CHANGED";
return PartialView("Part", vm);
}
View:
#model MvcApplication1.Controllers.HomeController.MYvm
#{
ViewBag.Title = "Home Page";
}
#using (Ajax.BeginForm("DA", "Home", new AjaxOptions() { UpdateTargetId = "cont", HttpMethod = "Get" }))
{
<div>
Id: #Html.EditorFor(model => model.id)
</div>
<div>
Name: #Html.EditorFor(model => model.name)
</div>
<input type="submit" value="SubmitForm" />
}
<div id="cont">
#{Html.RenderPartial("part", Model);}
</div>
Partial View
#model MvcApplication1.Controllers.HomeController.MYvm
#{
ViewBag.Title = "part";
}
<h2>part</h2>
#Model.name
Should I go with [previous SO answer]
No. That answer was for a different scenario where the question had a form with two submit buttons that wanted to do two different actions (and wasn't even the accepted answer to that question).
Your sample screenshot indicates that some javascript/jquery and ajax would solve the issue cleanly.
As you're new to MVC, try to keep it relatively simple. Break up the page into separate parts:
the containing page
the edit form
the list with remove
the edit/list work independently and should be written in a way that they could be put on any other page - the page is just there to contain them and doesn't do much else (obviously your real page will contain more, but add those parts as separate components as well).
1 Create actions for your list and edit forms that return partialviews - just the parts that are needed for that view (self-contained)
controller:
[HttpGet]
public ActionResult AddCost()
{
var model = new Cost();
return PartialView(model);
}
[HttpPost]
public void AddCost(Cost model)
{
if (ModelState.IsValid) {
db.SaveCost(model);...
}
}
form Views/Home/AddCost.cshtml:
#using (Ajax.BeginForm(...
{
<div class='editor-label'>#Html.LabelFor(model=>model.Description)</div>
...etc...
}
I'll leave you to set the Ajax.BeginForm properties. But make sure the on-success calls reloadCostList() (see below)
controller
public ActionResult CostList()
{
var model = db.loadCosts(); ...
return PartialView(model);
}
list, Views/Home/CostList.cshtml
#model IEnumerable<ViewModels.Cost>
<table>
<thead>
<tr>
<th>Cost Description</th>
...
<tbody>
#foreach (var cost in Model.Costs)
{
<tr data-id='#cost.Id'>
<td>#Html.DisplayFor(x=>cost.Description)</td>
...
<td><a href='#' class='remove-button'>Remove</a></td>
}
...
2 Create an action + view for the main page with placeholder for the form and calls the list partial-action, eg:
<div id="body">
<div id="formWrapper">
#Html.Action("AddCost")
</div>
<div id="listWrapper">
#Html.Action("ListView")
</div>
</div>
if you already load the data for the page, you can pass it directly to the partial, but there's no need:
#Html.Partial("ListView", Model.Costs)
this allows you to refresh the list via an ajax call, something like:
function reloadCostList() {
$(".listWrapper").load("Home/CostList");
}
(ideally, $.ajax and add some fancy UI to indicate loading)
3 Add a remove action to your controller
[HttpPost]
public void RemoveCost(int id)
{
}
4 Wire up the Remove link
$(function() {
$(".remove-button").click(function() {
var id = $(this).closest("tr").attr("id");
$.post("/Home/RemoveCost/" + id, null, function() {
$(".listWrapper").load("Home/CostList");
// or reloadCostList(); from above
// or:
//$(".listWrapper tr[id=" + id + "]").hide();
});
});
}
rather than re-load the entire list, you could just remove the row (add some fancy UI like fade-out...)

Redirecting from cshtml page

I want to redirect to a different view depending on the result of a dataset, but I keep getting returned to the page I am currently on, and can't work out why. I drop into the if statement the action gets called but once i return the view to the new page, it returns me back to the current page.
CSHTML page
#{
ViewBag.Title = "Search Results";
EnumerableRowCollection<DataRow> custs = ViewBag.Customers;
bool anyRows = custs.Any();
if(anyRows == false)
{
Html.Action("NoResults","Home");
}
// redirect to no search results view
}
Controller
public ActionResult NoResults()
{
return View("NoResults");
}
View I cant get too..
#{
ViewBag.Title = "NoResults";
}
<h2>NoResults</h2>
Change to this:
#{ Response.Redirect("~/HOME/NoResults");}
Would be safer to do this.
#{ Response.Redirect("~/Account/LogIn?returnUrl=Products");}
So the controller for that action runs as well, to populate any model the view needs.
Source
Redirect from a view to another view
Although as #Satpal mentioned, I do recommend you do the redirecting on your controller.
This clearly is a bad case of controller logic in a view. It would be better to do this in a controller and return the desired view.
[ChildActionOnly]
public ActionResult Results()
{
EnumerableRowCollection<DataRow> custs = ViewBag.Customers;
bool anyRows = custs.Any();
if(anyRows == false)
{
return View("NoResults");
}
else
{
return View("OtherView");
}
}
Modify NoResults.cshtml to a Partial.
And call this as a Partial view in the parent view
#Html.Partial("Results")
You might have to pass the Customer collection as a model to the Result action or in a ViewDataDictionary due to reasons explained here: Can't access ViewBag in a partial view in ASP.NET MVC3
The ChildActionOnly attribute will make sure you cannot go to this page by navigating and that this view must be rendered as a partial, thus by a parent view. cfr: Using ChildActionOnly in MVC
You can go to method of same controller..using this line , and if you want to pass some parameters to that action it can be done by writing inside ( new { } )..
Note:- you can add as many parameter as required.
#Html.ActionLink("MethodName", new { parameter = Model.parameter })

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