ASP.Net MVC How to pass data from view to controll

2019-01-01 11:13发布

问题:

I am completely new to ASP.Net and I am sure this is a very basic question I have a View in which there is a link to generate report but to be able to generate report I must ask the user to provide a suitable text name as well.

So far I have been able to pass data from server to view using Models passed from my controller to view, but I am not sure how to Pass data from view to my controller.

I just need to pass a string from view to controller in this case.

Any advice with example will be appreciated.

UPDATE

I understand I have to post the data back to server but how does that realize in the form of razorhtml code and controller?

回答1:

You can do it with ViewModels like how you passed data from your controller to view.

Assume you have a viewmodel like this

public class ReportViewModel
{
   public string Name { set;get;}
}

and in your GET Action,

public ActionResult Report()
{
  return View(new ReportViewModel());
}

and your view must be strongly typed to ReportViewModel

@model ReportViewModel
@using(Html.BeginForm())
{
  Report NAme : @Html.TextBoxFor(s=>s.Name)
  <input type=\"submit\" value=\"Generate report\" />
}

and in your HttpPost action method in your controller

[HttpPost]
public ActionResult Report(ReportViewModel model)
{
  //check for model.Name property value now
  //to do : Return something
}

OR Simply, you can do this without the POCO classes (Viewmodels)

@using(Html.BeginForm())
{
   <input type=\"text\" name=\"reportName\" />
   <input type=\"submit\" />
}

and in your HttpPost action, use a parameter with same name as the textbox name.

[HttpPost]
public ActionResult Report(string reportName)
{
  //check for reportName parameter value now
  //to do : Return something
}

EDIT : As per the comment

If you want to post to another controller, you may use this overload of the BeginForm method.

@using(Html.BeginForm(\"Report\",\"SomeOtherControllerName\"))
{
   <input type=\"text\" name=\"reportName\" />
   <input type=\"submit\" />
}

Passing data from action method to view ?

You can use the same view model, simply set the property values in your GET action method

public ActionResult Report()
{
  var vm = new ReportViewModel();
  vm.Name=\"SuperManReport\";
  return View(vm);
}

and in your view

@model ReportViewModel
<h2>@Model.Name</h2>
<p>Can have input field with value set in action method</p>
@using(Html.BeginForm())
{
  @Html.TextBoxFor(s=>s.Name)
  <input type=\"submit\" />
}


回答2:

In case you don\'t want/need to post:

@Html.ActionLink(\"link caption\", \"actionName\", new { Model.Page })  // view\'s controller
@Html.ActionLink(\"link caption\", \"actionName\", \"controllerName\", new { reportID = 1 }, null);

[HttpGet]
public ActionResult actionName(int reportID)
{

Note that the reportID in the new {} part matches reportID in the action parameters, you can add any number of parameters this way, but any more than 2 or 3 (some will argue always) you should be passing a model via a POST (as per other answer)

Edit: Added null for correct overload as pointed out in comments. There\'s a number of overloads and if you specify both action+controller, then you need both routeValues and htmlAttributes. Without the controller (just caption+action), only routeValues are needed but may be best practice to always specify both.



回答3:

<form action=\"myController/myAction\" method=\"POST\">
 <input type=\"text\" name=\"valueINeed\" />
 <input type=\"submit\" value=\"View Report\" />
</form> 

controller:

[HttpPost]
public ActionResult myAction(string valueINeed)
{
   //....
}