I am new to Asp.net MVC and have no idea as to how can i perform the search. Here's my requirement, please tell me how will you handle this :-
I need to have textbox where user can enter a search query or string. The user then clicks on a button or presses enter to submit it. The string needs to matched with a table's property name.
NOTE:- Querying the data and fetching the result isn't the main point here. All I need to know is how will you take the user input and pass it to a controller action or whatever for further processing. Just tell me how will you read the user input and where will you send it to search.
Asp.Net MVC uses standard HTTP verbs. For the html part, it's a normal html form that points to an url. Server side, that url will be routed to a controller/action which will handle the input and do what is needed.
Let's have a sample. You want to make a search form. First of all, it's a best practice to have search forms use the HTTP GET method instead of POST, so the search results can be bookmarked, linked, indexed, etc. I won't be using Html.BeginForm helper method to make things more clear.
<form method="get" action="#Url.Action("MyAction", "MyController")">
<label for="search">Search</label>
<input type="text" name="search" id="search" />
<button type="submit">Perform search</button>
</form>
That's all the html you need. Now you'll have a controller called "MyController" and the method will be something like this:
[HttpGet]
public ActionResult MyAction(string search)
{
//do whatever you need with the parameter,
//like using it as parameter in Linq to Entities or Linq to Sql, etc.
//Suppose your search result will be put in variable "result".
ViewData.Model = result;
return View();
}
Now the view called "MyAction" will be rendered, and the Model of that view will be your "result". Then you'll display it as you wish.
As always in an ASP.NET MVC application you start by defining a view model which will express the structure and requirements of your view. So far you have talked about a form containing a search input:
public class SearchViewModel
{
[DisplayName("search query *")]
[Required]
public string Query { get; set; }
}
then you write a controller:
public class HomeController : Controller
{
public ActionResult Index()
{
return View(new SearchViewModel());
}
[HttpPost]
public ActionResult Index(SearchViewModel model)
{
if (!ModelState.IsValid)
{
// There was a validation error => redisplay the view so
// that the user can fix it
return View(model);
}
// At this stage we know that the model is valid. The model.Query
// property will contain whatever value the user entered in the input
// So here you could search your datastore and return the results
// You haven't explained under what form you need the results so
// depending on that you could add other property to the view model
// which will store those results and populate it here
return View(model);
}
}
and finally a view:
#model SearchViewModel
#using (Html.BeginForm())
{
#Html.LabelFor(x => x.Query)
#Html.EditorFor(x => x.Query)
#Html.ValidationMessageFor(x => x.Query)
<button type="submit">Search</button>
}
This is the best way to do it.
Create a ViewModel
public class SearchViewModel
{
public string Query { get; set; }
}
Create a Controller
public class SearchController : Controller
{
[HttpPost]
public ActionResult Search(SearchViewModel model)
{
// perform search based on model.Query
// return a View with your Data.
}
}
Create the View
// in your view
#using (Html.BeginForm("Search", "SearchController"))
{
#Html.TextBox("Query")
<input type="submit" value="search" />
}
hope this helps
Related
Is it possible to somehow not clear a file input when posting back a viewmodel in ASP.NET MVC Core? I know the IFormFile doesn't include a member that designates the directory of the file selected.
So, if the view model gets sent back in the situation that ModelState is invalid the file input is cleared. Is it possible to avoid this without resorting entirely to clientside validation?
Here's an example controller and template to more adequately explain what I am saying
Controller:
namespace Example
{
public class CreateVm
{
public IFormFile aFile{ get; set; }
public string someText{get; set;}
}
public class ExampleController : Controller
{
public IActionResult Action()
{
var vm = new CreateVm();
return View(vm);
}
[HttpPost]
public IActionResult Create(CreateVm vm)
{
// Something invalidates the model state
// The supplied viewmodel isn't valid, send it back
if (!ModelState.IsValid)
return View(vm);
// Do stuff here
return RedirectToAction(nameof(Create));
}
}
}
Template:
#model Example.CreateVm
<form asp-action="Create" method="post" enctype="multipart/form-data">
<input asp-for="someText"/>
<input asp-for="aFile"/>
<input type="submit" value="Submit">
</form>
From what I understand it is not possible to get the file directory in javascript so I don't think that's a viable way to go about it either.
I want to show one TextBox. In that if give any input string and button clicked it should so like this
hai , what is ur name
[TextBox]
welcome,ur name is "xyz"
I am new in MVC. Please help me to do this.
View
#{
ViewBag.Title = "MyPage";
}
<h2>Mymethod</h2>
<h3>#ViewBag.Message</h3>
#Html.TextBox("Name")
<form method="post">
<input type="submit" value="Submit" name="btn" />
</form>
HomeController.cs
public ActionResult Mymethod()
{
ViewBag.Message = "Hello what is ur name ??? ";
return View();
}
There are many ways to do this to accomplish what you want. I will provide you with a simplistic approach, please modify and change it to fit in with your scenario.
I would normally recommend using a view model above any other way, for example using a single string value or using FormCollection or ViewBag. They can work but I prefer to use view models.
I answered a question on what view models are and what they are supposed to do, please read it:
What is ViewModel in MVC?
First you will create a view model that will handle your input data, like first name, last name, age, etc. You will then pass this view model through to the view. In your example I will only include name:
public class ViewModel
{
public string Name { get; set; }
}
In your Create action method you will instantiate this view model and pass it to the view. And when you click on the button to submit the form then the post action method will receive this view model as input parameter:
public ActionResult Create()
{
ViewModel model = new ViewModel();
return View(model);
}
[HttpPost]
public ActionResult Create(ViewModel model)
{
if (!ModelState.IsValid)
{
// If validation fails send the view model back to the view and fix any errors
return View(model);
}
// Do what you need to do here if there are no validation errors
// In your example we are posting back to the current view to
// display the welcome message
return View(model);
}
And then finally you view will look like this:
#model Project.Models.ViewModel
#using (Html.BeginForm())
{
#Html.TextBoxFor(m => m.Name)
<button type="submit">Submit</button>
if (!string.IsNullOrWhiteSpace(Model.Name))
{
<p>welcome, your name is #Model.Name</p>
}
}
Please spend some time reading through the many online tutorials on ASP.NET MVC.
Modify your current view to
#using(Html.BeginForm("ControllerName","Mymethod",FormMethod.Post))
{
<input type="submit" value="Submit" name="btn" />
}
Add another action method in your controller like this :
[HttpPost]
public ActionResult Mymethod(FormCollection form)
{
string Name = form["Name"];
Viewbag.Name = Name;
return View()
}
Then add view to this controller and write this into it.
Hi , Your Name is #Viewbag.Name
You should wrap your form in form tag. It is a form after all. So when you click submit, you are submitting the form.
<form method="post">
<h2>Mymethod</h2>
<h3>#ViewBag.Message</h3>
#Html.TextBox("Name")
#if (!String.IsNullOrEmpty(ViewBag.Name))
{
<h3>
welcome,ur name is #ViewBag.Name
</h3>
}
<input type="submit" value="Submit" name="btn" />
</form>
On the controller, you need to add HttpPost handler for your Mymethod action. This is where your web server is accepting the form you've submitted.
[HttpPost]
public ActionResult Mymethod(string name)
{
ViewBag.Message = "Hello what is ur name ???";
ViewBag.Name = name;
return View();
}
I am new to ASP.NET MVC. Is it possible to pass data from view to model in mvc? this question was asked in interview.Please anyone give me right answer.
Thanks in Advance
To pass data to controller through model you should wrap all the inputs (checkboxes, textboxes, radio etc.) with <form ...> tag. You could do it using HTML tag or with help of ASP.NET MVC helper #Html.BeginForm(...). Once you submit your form, all the input data will be sent to a controller action and mapped to a targeted model. Please see an example:
Model:
public class UserModel
{
public string FirstName { get; set; }
public string LastName { get; set; }
}
View:
#model UserModel
#using (Html.BeginForm("Search", "Events"))
{
#Html.TextBoxFor(m => m.FirstName)
#Html.TextBoxFor(m => m.LastName)
<input type="submit" value="Search" />
}
Controller:
public class EventsController: Controller
{
public ActionResult Search(UserModel model)
{
//do something
return View(); //return "Search" view to the user
//return View(model); //You can also return view with the model to the user
//return View("SpecificView"); //You can specify a concrete view name as well
}
}
No, we do not pass any information from view to model directly, both the view and model are different module. we can pass data, value or any information from view to model via controller.
I am making WCF service call using MyViewRequest view fields inside HttpPost ActionHandler. The goal is to show response using partial view, MyViewResponse
In brief I need to achieve these two items-
Disable load of partial view on first load.
Display Response (along with Request) after service call.
MyViewRequest.cshtml
#using (Html.BeginForm())
{
#Html.ValidationSummary(false)
//html code
}
</div>
<div id="dvResponse">
#Html.Partial("MyViewResponse");
</div>
Partial view: MyViewResponse.cshtml
#model MvcApplication3.Models.MyModel
#{
ViewBag.Title = "MyViewResponse";
}
<h2>MyView</h2>
#Html.Label(Model.MyName, "My Name")
This was pretty straight forward in Asp.Net using userControl, But stuck up here, How can we achieve this in MVC3.
I think the best way is to transfer your data using ViewModels. Let's assume you want to have an app something like stackoverflow where you have a question and user can post an answer and it will be shown after the post along with the question.
public class PostViewModel
{
public int ID { set;get;}
public string Text { set;get;}
public List<PostViewModel> Answers { set;get;}
public string NewAnswer { set;get;}
}
in your GET action, you show the question. Get the id from the url and get the Question details from your service/repositary.
public ActionResult Show(int id)
{
var post=new PostViewModel();
post=yourService.GetQuestionFromID(id);
post.Answers=yourService.GetAnswersFromQuestionID(id);
return View(post);
}
Assuming yourService.GetQuestionFromID method returns an object of PostViewModel with the propety values filled. The data can be fetched from your database or via a WCF service call. It is up to you. Also yourService.GetAnswersFromQuestionID method returns a list of PostViewModel to represent the Answers for that question. You may put both these into a single method called GetQuestionWithAnswers. I wrote 2 methods to make it more clear.
Now in your Show view
#model PostViewModel
#Html.LabelFor(x=>x.Text);
#using(Html.Beginform())
{
#Html.HiddenFor(x=>x.ID);
#Html.TextBoxFor(x=>x.NewAnswer)
<input type="submit" />
}
<h3>Answers</h3>
#if(Model.Answers!=null)
{
#Html.Partial("Responses",Model.Answers)
}
And your Partial view will be strongly typed to a collection of PostViewModel
#model List<PostViewModel>
#foreach(var item in Model)
{
<div> #item.Text </div>
}
Handling the postback is simple (HttpPost)
[HttpPost]
public ActionResult Show(PostViewModel model)
{
if(ModelState.IsValid)
{
//get your data from model.NewAnswer property and save to your data base
//or call WCF method to save it.
//After saving, Let's redirect to the GET action (PRG pattern)
return RedirectToAction("Show",new { #id=model.ID});
}
}
I'm making an invite system where a user when registering to the site can specify a user who refereed them.
There's also a way for existing users to send invites. The friend would recieve a link:
http://www.foo.com/account/register?referal=sandyUser216
How can I get that value sandyUser216 and place it as the value inside of a text input box?
I'm using C# and MVC3.
Check the Request.QueryString.
<input type="text" value="#Request.QueryString["referal"]" />
Or put it in a Model property rather than having it in the view.
As always in an ASP.NET MVC application you start by writing a view model that will represent the information contained in your view:
public class RegisterViewModel
{
[Required]
public string Referal { get; set; }
}
then you write controller actions for respectively showing the registration form and processing it:
public ActionResult Register(RegisterViewModel model)
{
return View(model);
}
[HttpPost]
[ActionName("Register")]
public ActionResult ProcessRegistration(RegisterViewModel model)
{
if (!ModelState.IsValid)
{
return View(model);
}
// TODO: perform the registration
return RedirectToAction("success");
}
and finally you write the corresponding strongly typed view:
#model RegisterViewModel
#using (Html.BeginForm())
{
#Html.LabelFor(x => x.Referal)
#Html.EditorFor(x => x.Referal)
#Html.ValidationMessageFor(x => x.Referal)
<button type="submit">Register</button>
}
Now all that's left is to simply navigate to /account/register?referal=sandyUser216.
And you have accomplished the whole MVC pattern. Should you skip any of those 3 letters it means that you are doing ASP.NET MVC incorrectly.