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Home/ Questions/Q 6049497
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T07:35:01+00:00 2026-05-23T07:35:01+00:00

I have this code in the View: <script src=../../Scripts/jquery.form.js type=text/javascript></script> <script src=../../Scripts/jquery-1.5.1.js type=text/javascript></script> <script

  • 0

I have this code in the View:

<script src="../../Scripts/jquery.form.js" type="text/javascript"></script>
<script src="../../Scripts/jquery-1.5.1.js" type="text/javascript"></script>

<script type="text/javascript">
$(document).ready(function () {
    $('#myForm').ajaxSubmit();
});

@using (Html.BeginForm("PostData","Home",FormMethod.Post)) {

 <input type="text" name="name" />
 <input type="submit" value="submit" />
}

Then in the controller I have this method:

 [HttpPost]
    public void PostData(string name)
    {
        //do smoething with name
    }

The problem is that I get redirected the url /home/PostData when submitting the form.

Anyone got any suggestions?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T07:35:01+00:00Added an answer on May 23, 2026 at 7:35 am

    ajaxSubmit() submits the form immediately. You need ajaxForm() in order to AJAXify an existing form:

    $(document).ready(function () {
        $('#myForm').ajaxForm();
    });
    

    once a form has been AJAXified you could later force its submission with $('#myForm').ajaxSubmit(); or simply leave it to the user to press the submit button.

    Also the way you have defined your form it doesn’t seem to have an id. So your $('#myForm') selector is unlikely to return anything. You might want to assign an id to your form:

    @using (Html.BeginForm("PostData", "Home", FormMethod.Post, new { id = "myForm" })) 
    {
        <input type="text" name="name" />
        <input type="submit" value="submit" />
    }
    

    or if you don’t want to assign unique id to the form you could AJAXify all forms on the current view by adapting your jQuery selector:

    $(document).ready(function () {
        $('form').ajaxForm();
    });
    
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