CodeIgniter AJAX, doesn't update/receive new data in database - php

Question

I have a problem with updating database using AJAX and CodeIgniter. When someone posts AJAX form I retrieve data about user from database this way:
class MY_Controller extends CI_Controller
{
public $memberData;
public function __construct()
{
parent::__construct();
$this->memberData= $this->membermodel->getmemberData();
}
}
(Every controller extend MY_Controller, not CI_Controller).
Then I operate on user data and insert it to database. The problem is that if I send AJAX post really fast (few at same time), few inserted rows are identical (except auto-increment row ID). It looks like CodeIgniter did not receive new user data from database (or don't update it before), and I operate on old one.
I send AJAX like this:
$("#form_id").submit(function(event)
{
$form = $(this);
$.post($form.attr('action'), $(this).serialize(), function(HTML)
{
//do something
});
return false;
});
then I operate on something like this:
$CI =& get_instance();
$CI->load->model('membermodel', 'member');
$variab['value3'] = $CI->memberData->member_value3 + 1; //this is the line that need to be new on every call, but it doesnt
$variab['result'] = $this->calculatedata($variab['value3']);
$parameters = array(
'member_value3' => $variab['value3']
);
//update that variable to database, so it should have new value new on next call
$CI->member->updateinfo($parameters);
return $variab['value3'];
At the end I get that value3, base on it my whole script and insert last query to database. Unfortunately like I said if I send many POST requests in the same time that value is constant.

Answer 1

I use this, it should work out of clipboard, I found it a while ago on SO don't remember the source. (will search for it) source.
Please do read comments and do debug using console.
var r;
$("#form_id").submit(function(event){
if (r) {
r.abort(); //abort all previous requests
}
var $form = $(this);
var $inputs = $form.find("input, textarea, select, button"); //serialize all elements in form
var serializedData = $form.serialize(); //serialize data (prep data)
$inputs.prop("disabled", true); //disable inputs
r = $.ajax({
url: $form.attr('action'), //whatever.php
type: "post",
data: serializedData
});
// Callback handler that will be called on success
r.done(function (response, textStatus, jqXHR){
console.log("Form works: " + response);
});
// Callback handler that will be called on failure
r.fail(function (jqXHR, textStatus, errorThrown){
console.error(
"Error: "+
textStatus, errorThrown
);
});
r.always(function () {
$inputs.prop("disabled", false); //enable inputs
});
// Prevent default posting of form
event.preventDefault();
});
Regard to comment: But if somebody change this JS code he still will be able to bug my whole script, I have to get it secured.
Please consider following order of developing.
create form using HTML
create php file that accepts this form and VALIDATES it; debug all possible inputs etc.
create AJAX part
style using CSS, show errors etc.
This way you will never (rarely) have problems in matter you just did.