I want to validate a form without refreshing the page using the .post() jQuery method. I use codeigniter for validation . Here is the jQuery code:

$(document).ready(function(){

$(".form_errors").hide();

$("#send").on("click",function(){ //the submit button has the id="send"

$(".form_errors").hide();  //these are <p> for each input to show the error

var user=$("input.box");

var data={};

var names=$("input.box").attr("name");

for(i=0;i<user.length;i++){

name=names[i];

value=user[i].val();

data.name=value;

}

$.post('ksite/account',

data,

function(result){

$("div.answer").html(result);

for(i=0;i<user.length;i++){

error_message=<?php echo form_error("?>names[i]<?php ");?>;

$("p#error_"+names[i]+".form_errors").html(error_message).show();

}

}

});

});

form_error is a CodeIgniter function.(I suppose someone who used ci is familiar with).

The form :

<p id="error_user" class="form_errors"></p>

<input type="text" class="box" name="user">

<p id="error_password" class="form_errors"></p>

<input type="password" class="box" name="password">

<input type="submit" id="send">

Is the form tag neccessary ? And if yes,do i have to mention action and method ?

Do I have to specify the type of the response?

And in ksite/account I do :

......

if(!this->form_validation->run(''account")){

    echo "The account couldn't be made";

}

else

{

  echo "The account was successfully created  ";

}

P.S.Although you may not be familiar with codeigniter,I would appreciate if someone coulf tell me if the code is correct and what improvements could be made .

Thanks a lot !