如何将数据从视图传递到控制器

<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)
{
//....
}

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" />
}

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.