MVC View to controller Passing

Question

I am starting up with mvc and designing a simple login procedure. I have view with login screen with two inputs username and password. but apparently not able to get how can i pass the values of inputs from my view to controller i am using razor. here are my snippets.

<table>
    <tr>
        <td>
            UserName:
        </td>
        <td>
            @Html.TextBox("userName")
        </td>
    </tr>
        <tr>
        <td>
            Password
        </td>
        <td>
            @Html.Password("Password")
        </td>
    </tr>
    <tr>
        <td colspan="2">
            @Html.ActionLink("login", "SignIn")
        </td>
    </tr>
</table>

and my controller looks like this.( i am able to redirect to controller using action link just fine.just about passing values.)

public ActionResult SignIn()
{
    //string userName = Request["userName"];
    return View("Home");
}

Akash KC · Accepted Answer · 2012-09-27 10:01:09Z

2

You can enclosed above html stuff inside form container where you declared form submission method as POST.

@using (Html.BeginForm("SignIn", "Controller", FormMethod.Post, new { id = "form1" }))
{
    <table>
    <tr>
        <td>
            UserName:
        </td>
        <td>
            @Html.TextBox("userName")
        </td>
    </tr>
        <tr>
        <td>
            Password
        </td>
        <td>
            @Html.Password("Password")
        </td>
    </tr>
    <tr>
        <td colspan="2">
            <input type="submit" value="login" name="login" />
        </td>
    </tr>
</table>
}

Then, you can put Post Action in your controller:

[HttpPost]
public ActionResult SignIn(FormCollection frmc)
{
    /// Extracting the value from FormCollection
    string name = frmc["userName"];
    string pwd = frmc["Password"];
    return View("Home");
}

edited Sep 27, 2012 at 10:01

answered Sep 27, 2012 at 9:55

Akash KC

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2 Comments

Mandar Jogalekar Over a year ago

perfect! that brings me to question.is that the best practice? or is there something better?

Wouter Simons Over a year ago

I would make the view with an @model VmLoginParams and make the controller post action take public ActionResult SignIn(VmLoginParams params) In the Html code you can then do @Html.EditorFor(m => m.UserName) @Html.ValidationMessageFor(m => m.UserName) and in the controller check for ModelState.IsValid: if (!ModelState.IsValid) return View(params) (See also: mikesdotnetting.com/Article/188/…)

Azargoth · Accepted Answer · 2012-09-27 09:55:04Z

0

Wrap your table in Form:

    @using (Html.BeginForm("SignIn", "controllerName", FormMethod.POST))
{
<table>
...
</table>
<input type="submit" value="Sign in" />
}

And in controller write:

[HttpPost]
public ActionResult SignIn(string userName, string Password)
{
 //sign in and redirect to home page
}

answered Sep 27, 2012 at 9:55

Azargoth

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Comments

Ranjith Kumar Nagiri · Accepted Answer · 2014-02-01 18:13:46Z

0

View:

@using (Html.BeginForm("SignIn", "Controller", FormMethod.Post, new { id = "form1" }))
{
<table>
<tr>
    <td>
        UserName:
    </td>
    <td>
        @Html.TextBox("userName")
    </td>
</tr>
    <tr>
    <td>
        Password
    </td>
    <td>
        @Html.Password("Password")
    </td>
</tr>
<tr>
    <td colspan="2">
        <input type="submit" value="login" name="login" />
    </td>
</tr>
</table>
}

Model:

public string userName{get;set;}
public string Password{get;set;}

Controller:

[HttpPost]
 public ActionResult SignIn(Model obj)
 {
  //sign in and redirect to home page
   string userName = obj.username;
   string password = obj.password;
 }

it may be help full to you.

edited Feb 1, 2014 at 18:13

answered Jan 31, 2014 at 20:00

Ranjith Kumar Nagiri

8613 gold badges18 silver badges42 bronze badges

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