asp.net MVC 4 submit from with ajax call

Question

I have this in the controller:

[HttpPost]
    public ActionResult Create(Student student)
    { //somecode..

and i want to submit this from:

<form method="post" action="/student/create"> 
<!-- the from contents-->

how to submit this from using Ajax call i need the JQuery ajax call that lets this form submitted.

and i want to make sure about datatype , thanks

Umesh Sehta · Accepted Answer · 2013-11-12 08:23:28Z

18

try this

var form = $('#formId');
$.ajax({
  cache: false,
  async: true,
  type: "POST",
  url: form.attr('action'),
  data: form.serialize(),
  success: function (data) {
    alert(data);
 }
});

answered Nov 12, 2013 at 8:23

Umesh Sehta

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3 Comments

Muath Over a year ago

form = document.forms[2]--- when i use form.serialize() , i have type error [serialize is not function] JQuery already included.. how to solve it?

Umar Abbas Over a year ago

can you please tell me that how can we trigger default mvc validation before submitting form?

Umesh Sehta Over a year ago

@UmarAbbas you can try like this:- jsfiddle.net/cfwx3Ld5 and use razor form instread of html

arulmr · Accepted Answer · 2014-07-20 09:53:16Z

5

Use this, assuming you are using razor views:

@using (Ajax.BeginForm(new AjaxOptions(){
HttpMethod = "POST",
    Url = "your controller",
    OnComplete = "some client event"
})
{
    <fieldset>
        <legend>This is a demo form.</legend>
        @Html.LabelFor(model => model.Name)
        @Html.TextBoxFor(model => model.Name)

        <input type="submit" value="Save" />
    </fieldset>
}

edited Jul 20, 2014 at 9:53

arulmr

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answered Nov 12, 2013 at 8:30

Besher

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1 Comment

mokumaxCraig Over a year ago

You're missing a end parens in your using. S/B: OnComplete = "some client event"}))

Dimitar Dimitrov · Accepted Answer · 2013-11-12 08:22:44Z

1

Well it would look something like (that's without looking at your view bindings):

// serialize your form into JSON - maybe you have a different method of doing it
var serializedForm = $("#YourFormId").serialize();

// post it to the server
$.post("/student/create", serializedForm)
    .done(function (response) { 
       // it's done
    })
    .fail(function (xhr, status, error) { 
       // something bad happened
    });

answered Nov 12, 2013 at 8:22

Dimitar Dimitrov

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