CodeIgniter : Setting up jQuery and post data to Controller using jQuery

Question

I added

<script type="text/javasript" src="<?=base_url()?>js/jquery-1.7.2.min.js"></script>

in application/views/templatess/header.php and to post data to Controller using jQuery I wrote

<script>

$(document).ready(function(){

    $('#submit').click(function() {
        $.ajax({
            type : 'POST',
            url : '/projects/create',
            data: {
                pro_name : $('#pro_name').val()
            },
            success:function (data) {
                $("#log_msg").html(data);
            }          
        });
    });

});
</script>

in my view file: application/views/project/create.php and the controller exists in application/controllers/projects.php

In my view named create.php the complete code is:

<label for="pro_name">Project Name</label>
<input type="input" id="pro_name" name="pro_name" />
<br />
<input type="submit" id="submit" name="submit" value="Create" />

<script>

$(document).ready(function(){

    $('#submit').click(function() {
        $.ajax({
            type : 'POST',
            url : '/projects/create',
            data: {
                email : $('#pro_name').val()
            },
            success:function (data) {
                $("#log_msg").html(data);
            }          
        });
    });

});
</script>

and my controller named projects contains

class Projects extends CI_Controller {
    public function __construct()
    {
        parent::__construct();
        $this->load->model('projects_model');
        $this->load->helper('url');
        $this->load->helper('html');

        $this->load->helper('form');
        $this->load->library('form_validation');
    }

    public function create()
    {
        echo $this->input->post('pro_name');

        $data['title'] = 'SPARCS | Create Project';

        $this->form_validation->set_rules('pro_name', 'Name', 'required');
        $this->form_validation->set_rules('pro_client', 'Client', 'required');
        $this->form_validation->set_rules('pro_loc_city', 'City', 'required');
        $this->form_validation->set_rules('pro_loc_state', 'State', 'required');
        $this->form_validation->set_rules('pro_size', 'Project Size', 'required');      
        $this->form_validation->set_rules('pro_desc', 'Description', 'required');

        if ($this->form_validation->run() === FALSE)
        {
            $this->load->view('templates/header', $data);   
            $this->load->view('projects/create');
            $this->load->view('templates/footer');

        }
        else
        {
            // If project name already exists?

            $query = $this->projects_model->if_exists($this->input->post('pro_name'));

            if ( sizeof($query) == 0) {

                //$this->projects_model->set_project();             
                $data['message'] = 'Add successfully';

            } else {

                $data['message'] = 'Project name already exists';
            }

            $this->load->view('projects/log_message', $data);
        }
    }
}

But browser says $ is not defined

Now please let me know how to setup jQuery in CodeIgniter and what is the correct way to pass data to the controller and also show response send back from the controller

Where in the browser does it say $ is not defined? Are you sure your jQuery path is correct? — Ayman Safadi
– Ayman Safadi, Commented May 27, 2012 at 5:01
In firebug, yes jquery is located at root/js/jquery.js I saw view source and when clicked and it is correct and if i clicked it show the js file — PHP Ferrari
– PHP Ferrari, Commented May 27, 2012 at 5:09
I recommend that you change the url in your ajax request from /projects/create to <?= site_url('projects/create'); ?> — Ayush
– Ayush, Commented May 27, 2012 at 5:25
I also follow stackoverflow.com/questions/5051059/… but same error shows $ is not defined — PHP Ferrari
– PHP Ferrari, Commented May 27, 2012 at 6:27
Okay I fix it we should avoid to use <?= instead we should use <?php echo "your_script_here";?> So I replace <?=base_url()?> with <?php echo base_url();?> — PHP Ferrari
– PHP Ferrari, Commented May 27, 2012 at 6:36

Ayman Safadi · Accepted Answer · 2012-05-27 05:19:31Z

0

Is it possible that somewhere in your code, jQuery is being set into "safe mode"?

Try wrapping your logic in an anonymous function, like this:

(function($) {
    // insert logic here
})(jQuery)

Whatever is in this function is guaranteed to work with $ as a shortcut for jQuery.

answered May 27, 2012 at 5:19

Ayman Safadi

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