ajax function not going to php codeigniter controller

Question

I have a ajax function written which is posting different numbers. Here is the ajax function.

self.giveCashtoChild = function(){
            $.ajax({
                type: 'POST',
                url: BASEURL + '/index.php/main/addUserChildrenCash'+"/"+self.selectedchild(),
                contentType: 'application/json; charset=utf-8'
            })
            .done(function() {

            })
            .fail(function(xhr, status, error) {
                alert(status);
            })
            .always(function(data){                 
            });
        }

self.selectedchild() has the value of 2 , so Basically the url is addUserChildrenCash/2 but then it does not go to the codeigniter controller and change the page. Here is the controller function.

public function addUserChildrenCash($childID){
            if (!$this->session->userdata('user_id')){
                redirect('main'); // the user is not logged in, redirect them!
            }

                $userid= $this->session->userdata('user_id');
                $this->load->model('main_page');
                $childname =  $this->main_page->getChildName($childID, $userid);

                $data = array(
                    'name' => $childname['children_name']
                );
                $this->load->view('header2_view');
                $this->load->view('add_user_children_cash_view' , $data);
                $this->load->view('footer_view');

        }

Are you using CodeIgniter's built-in CSRF protection? If so, then your ajax data must include the value of the hidden CSRF token field. This is the part that gets submitted with the standard form action but is missing when you do ajax. — Sparky
– Sparky, Commented Mar 25, 2016 at 14:26
Check for CSRF in your CodeIgniter configuration file. See: codeigniter.com/user_guide/general/… — Sparky
– Sparky, Commented Mar 25, 2016 at 14:32
How can I get rid of that ? $config['csrf_protection'] = FALSE; $config['csrf_token_name'] = 'csrf_test_name'; $config['csrf_cookie_name'] = 'csrf_cookie_name'; $config['csrf_expire'] = 7200; $config['csrf_regenerate'] = TRUE; $config['csrf_exclude_uris'] = array(); — XAF
– XAF, Commented Mar 25, 2016 at 14:34
Get rid of what? Looks like you already have it deactivated... $config['csrf_protection'] = FALSE Although I strongly suggest you learn about what it means and learn how to use it. — Sparky
– Sparky, Commented Mar 25, 2016 at 14:41

Abdulla Nilam · Accepted Answer · 2016-03-25 14:45:38Z

1

You define ajax as POST but you sending it via GET

 type: 'POST',
 url: BASEURL + '/index.php/main/addUserChildrenCash'+"/"+self.selectedchild(),

So your code should be

In Ajax

var id = self.selectedchild(); # Assign data to here
$.ajax(
    {

        type:"post",
        url: "<?php echo base_url(); ?>index.php/main/addUserChildrenCash", 
        data:{ id:id},
        success:function()
        {

        }
        error:function()
        {

        }
        always:function(){

        }
    });

In controller

public function addUserChildrenCash()
{
    $id = $this->input->post("id");
    echo $id
}

answered Mar 25, 2016 at 14:45

Abdulla Nilam

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5 Comments

XAF Over a year ago

Tried it , not sure why its not working, I do get all variables.

Abdulla Nilam Over a year ago

console.log(id) and check console

Abdulla Nilam Over a year ago

did you change base url to this <?php echo base_url(); ?>??

XAF Over a year ago

console.log(id) value comes to 2 and yes i did that second one as well

Abdulla Nilam Over a year ago

oohh Happy face at end :)

Ashkan Mobayen Khiabani · Accepted Answer · 2016-03-25 14:10:01Z

0

self.giveCashtoChild = function(){
            var id = self.selectedchild();
            $.ajax({
                type: 'POST',
                url: BASEURL + '/index.php/main/addUserChildrenCash/"+id,
                contentType: 'application/json; charset=utf-8'
            })
            .done(function() {

            })
            .fail(function(xhr, status, error) {
                alert(status);
            })
            .always(function(data){                 
            });
        }

answered Mar 25, 2016 at 14:10

Ashkan Mobayen Khiabani

34.4k35 gold badges112 silver badges187 bronze badges

2 Comments

Ashkan Mobayen Khiabani Over a year ago

please use an alert(id); after var id = self.selectedchild(); to see if it has the correct value

XAF Over a year ago

Yup has the right value of 2, but its not sending it to the controller

Malvolio · Accepted Answer · 2016-03-25 14:53:41Z

0

In your codeigniter use case I would pass your ajax parameter like this:

 $.ajax({
   type: 'post',
   url: BASEURL +'/index.php/main/addUserChildrenCash',
   data: { 'childID': self.selectedchild() },
   })
  ...

In the codeigniter controller I would receive the parameter like this:

public function addUserChildrenCash() {
    $childID = $this->input->post("childID");
    ...
    }

You should also check wether your BASEURL has the correct value assigned. Hope this helps.

edited Mar 25, 2016 at 14:53

answered Mar 25, 2016 at 14:25

Malvolio

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1 Comment

XAF Over a year ago

why have CHildID inside public function addUserChildrenCash() ?? and yea did not work :(

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