blocks|key|170831|text|<form+action="myController/myAction"+method="POST">
+<input+type="text"+name="valueINeed"+/>
+<input+type="submit"+value="View+Report"+/>
</form>+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|170832|控制器：|unstyled|170833|[HttpPost]
public+ActionResult+myAction(string+valueINeed)
{
+++//....
}|170834|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>&lt;form action="myController/myAction" method="POST"&gt;
 &lt;input type="text" name="valueINeed" /&gt;
 &lt;input type="submit" value="View Report" /&gt;
&lt;/form&gt; 
</code></pre>

<p>controller:</p>

<pre><code>[HttpPost]
public ActionResult myAction(string valueINeed)
{
   //....
}
</code></pre>


blocks|key|169107|text|您可以使用ViewModels完成此操作，就像您如何将数据从控制器传递到视图一样。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|169108|假设您有一个这样的视图模型|169109|public+class+ReportViewModel
{
+++public+string+Name+{+set;get;}
}|code-block|syntax|javascript|169110|在你的GET+Action中，|169111|public+ActionResult+Report()
{
++return+View(new+ReportViewModel());
}|169112|并且您的视图必须强类型化为ReportViewModel|offset|length|style|CODE|169113|@model+ReportViewModel
@using(Html.BeginForm())
{
++Report+NAme+:+@Html.TextBoxFor(s=>s.Name)
++<input+type="submit"+value="Generate+report"+/>
}|169114|在控制器中的HttpPost操作方法中|BOLD|169115|[HttpPost]
public+ActionResult+Report(ReportViewModel+model)
{
++//check+for+model.Name+property+value+now
++//to+do+:+Return+something
}|169116|或简单地说，您可以在没有POCO类(Viewmodels)的情况下做到这一点。|169117|@using(Html.BeginForm())
{
+++<input+type="text"+name="reportName"+/>
+++<input+type="submit"+/>
}|169118|在HttpPost操作中，使用与textbox名称同名的参数。|169119|[HttpPost]
public+ActionResult+Report(string+reportName)
{
++//check+for+reportName+parameter+value+now
++//to+do+:+Return+something
}|169120|编辑:根据评论编辑|169121|如果你想post到另一个控制器，你可以使用BeginForm方法的this+overload。|169122|@using(Html.BeginForm("Report","SomeOtherControllerName"))
{
+++<input+type="text"+name="reportName"+/>
+++<input+type="submit"+/>
}|169123|将数据从操作方法传递到视图？|169124|您可以使用相同的视图模型，只需在GET操作方法中设置属性值|169125|public+ActionResult+Report()
{
++var+vm+=+new+ReportViewModel();
++vm.Name="SuperManReport";
++return+View(vm);
}|169126|在你看来|169127|@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"+/>
}|169128|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/dd492590%2528v=vs.118%2529.aspx^0|0|0|0|0|0|D|F|0|0|6|8|0|0|0|1|0|0|0|0|0|9|0|X|D|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1W|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1X|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|1Z|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|20|8|@$O|21|P|22|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|F|7|23|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|24|8|@$O|25|P|26|Q|W]]|9|@]|A|$]]|$1|X|3|Y|5|F|7|27|8|@]|9|@]|A|$G|H]]|$1|Z|3|10|5|6|7|28|8|@$O|29|P|2A|Q|W]]|9|@]|A|$]]|$1|11|3|12|5|F|7|2B|8|@]|9|@]|A|$G|H]]|$1|13|3|14|5|6|7|2C|8|@]|9|@]|A|$]]|$1|15|3|16|5|F|7|2D|8|@]|9|@]|A|$G|H]]|$1|17|3|18|5|6|7|2E|8|@$O|2F|P|2G|Q|W]]|9|@]|A|$]]|$1|19|3|1A|5|6|7|2H|8|@]|9|@$O|2I|P|2J|1|2K]]|A|$]]|$1|1B|3|1C|5|F|7|2L|8|@]|9|@]|A|$G|H]]|$1|1D|3|1E|5|6|7|2M|8|@]|9|@]|A|$]]|$1|1F|3|1G|5|6|7|2N|8|@]|9|@]|A|$]]|$1|1H|3|1I|5|F|7|2O|8|@]|9|@]|A|$G|H]]|$1|1J|3|1K|5|6|7|2P|8|@]|9|@]|A|$]]|$1|1L|3|1M|5|F|7|2Q|8|@]|9|@]|A|$G|H]]|$1|1N|3|-4|5|6|7|2R|8|@]|9|@]|A|$]]]|1O|$1P|$5|1Q|1R|1S|A|$1T|1U]]]]

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

<p>Assume you have a viewmodel like this</p>

<pre><code>public class ReportViewModel
{
   public string Name { set;get;}
}
</code></pre>

<p>and in your GET Action,</p>

<pre><code>public ActionResult Report()
{
  return View(new ReportViewModel());
}
</code></pre>

<p>and your view must be strongly typed to <code>ReportViewModel</code></p>

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

<p>and in your <strong>HttpPost</strong> action method in your controller</p>

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

<p><strong>OR</strong> Simply, you can do this without the POCO classes (Viewmodels)</p>

<pre><code>@using(Html.BeginForm())
{
   &lt;input type="text" name="reportName" /&gt;
   &lt;input type="submit" /&gt;
}
</code></pre>

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

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

<p><strong>EDIT :</strong> <em>As per the comment</em></p>

<p>If you want to post to another controller, you may use <a href="http://msdn.microsoft.com/en-us/library/dd492590%28v=vs.118%29.aspx">this overload</a> of the BeginForm method.</p>

<pre><code>@using(Html.BeginForm("Report","SomeOtherControllerName"))
{
   &lt;input type="text" name="reportName" /&gt;
   &lt;input type="submit" /&gt;
}
</code></pre>

<h3>Passing data from action method to view ?</h3>

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

<pre><code>public ActionResult Report()
{
  var vm = new ReportViewModel();
  vm.Name="SuperManReport";
  return View(vm);
}
</code></pre>

<p>and in your view</p>

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


blocks|key|2980163|text|如果你不想/不需要发帖：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2980164|@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)
{|code-block|syntax|javascript|2980165|请注意，新{}部分中的reportID与操作参数中的reportID匹配，您可以通过这种方式添加任意数量的参数，但只要超过2或3个(有些人总是会争论)，您就应该通过POST传递模型(根据其他答案)|2980166|编辑：为注释中指出的正确重载添加null。有许多重载，如果同时指定action%2Bcontroller，则需要同时指定routeValues和htmlAttributes。如果没有控制器(只需要caption%2Baction)，则只需要routeValues，但最好总是同时指定两者。|offset|length|style|BOLD|2980167|entityMap^0|0|0|0|0|2|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

<p>In case you don't want/need to post:</p>

<pre><code>@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)
{
</code></pre>

<p>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)</p>

<p><strong>Edit</strong>: 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.</p>


<p>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.</p>

<p>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.</p>

<p>I just need to pass a string from view to controller in this case.</p>

<p>Any advice with example will be appreciated.</p>

<p><strong>UPDATE</strong></p>

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


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

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我是一个全新的ASP.Net，我相信这是一个非常基本的问题，我有一个视图，其中有一个链接来生成报告，但为了能够生成报告，我必须要求用户提供一个合适的文本名称以及。到目前为止，我已经能够使用从控制器传递到视图的模型将数据从服务器传递到视图，但我不确定如何将数据从视图传递到控制器。在这种情况下，我只需要将一个字符串从视图传...

问ASP.Net MVC如何将数据从视图传递到控制器
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