how to get values from database in codeigniter based on select box value without refreshing page?

Question

Unable to get value from database in codeigniter. I tried to fetch data based on select box value(menu_code) without refreshing page using ajax but I got result undefined.

This my controller's code : login.php

public function get_menu_rights()
{
    if (isset($_POST['name']))
    {       
        $root_id = $this->input->post('menu_root_id');

        $data['res'] = $this->login_model->get_menu_check($root_id);
        // print_r($data['res']);
        echo json_encode($data);
        //$this->load->view('pages/role_rights',$data);     
    }
}

Below is my model code login_model.php

public function get_menu_check($root_id)
{       
    $this->db->select('menu_code,menu_name');       
    $this->db->from('create_menu as C1');   
    $this->db->where('C1.menu_root_id',$root_id);       
    $this->db->order_by('menu_code');

    return $this->db->get()->result_array();
}

This is my view code role_rights.php

<form action="<?php echo base_url('login/get_menu_rights');?>" method="post">
    <?php
     print"<select class=\"form-control\" name=\"menu_root_id\" onchange=\"javascript:__doPostBack();\" id=\"menu_root_id\">"; ?>                                       <option value="select">select</option>
                        <?php foreach($result as $res) { ?>                             
                    <option value="<?php echo $res->menu_code; ?>">
                    <?php echo $res->menu_name.'-'.$res->menu_code; ?>
                    </option>
                <?php } ?>
            </select>
            </form>
        </div>
    <script src='https://code.jquery.com/jquery-2.1.3.min.js'></script>

<script type="text/javascript">
        function __doPostBack()
        {
            var name = document.getElementById('menu_root_id').value; 
            var dataString='name='+ name;
            $.ajax({
                type:"post",
                url:"<?php echo base_url('login/get_menu_rights'); ?>",
                data:dataString,
                cache:false,
                dataType: 'json',                  
              success: function(data)        
              {
                var id = data[0];             
                var vname = data[1]; 
                $('#output').html("<b>menu_code: </b>"+id+"<b> menu_name: </b>"+vname); 
              }
            });
            return false;
        }
        </script>
</div>

<div id="output"></div>

$data['res'] = ... You're putting the result row within 'res'. So you need to access it the same way. data.res.menu_code — Patrick Q
– Patrick Q, Commented Jul 19, 2018 at 14:13

Pradeep · Accepted Answer · 2018-07-19 14:48:35Z

1

Hope this will help you :

Replace

$root_id = $this->input->post('menu_root_id');

with

$root_id = $this->input->post('name');

Your controller's get_menu_rights method should be like this :

public function get_menu_rights()
{
  $root_id = $this->input->post('name');
  if(! empty($root_id))
  {
    $data = $this->login_model->get_menu_check($root_id);
    // print_r($data);
    echo json_encode($data);
    exit;   
   }
}

Your ajax success function should be like this :

success: function(data)        
{
  var html = '';
  $.each(data,function(k,v){
    alert(v);
    html += "<b>menu_code: </b>"+v.menu_code+"<b> menu_name: </b>"+v.menu_name
  });
  $('#output').html(html); 
}

edited Jul 19, 2018 at 14:48

answered Jul 19, 2018 at 13:54

Pradeep

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11 Comments

sunshine Over a year ago

but still it displays like this (menu_code: undefined menu_name: undefined). i passing my selected box value into ajax and getting response from ajax.So,i think i maked mistake in ajax..but i donno where...

Pradeep Over a year ago

where and on which file/line u r getting that error?

sunshine Over a year ago

i got error on <div id="output"></div> i think i'm not properly getting response using ajax from controller

sunshine Over a year ago

i got all values from db ...response in console successfully bro but how to display in my view file?{res: Array(2)} res : Array(2) 0 : {menu_code: "2", menu_name: "Plant"} 1 : {menu_code: "3", menu_name: "Line"} length : 2 proto : Array(0) proto : Object

Pradeep Over a year ago

ok , parse this data in your ajax, since it contains array data so you have to loop it see my updated answer for success function

|

Ashitaka · Accepted Answer · 2018-07-19 14:46:23Z

1

There are a few things I noticed

$data is an undefined array & you are settings the result array returned by the model function to it's 'res' key
dataString is not a json neither it's a js array that you are sending
since you used json_encode, you need to use JSON.parse(data) in the ajax success
if you do have the result in $data['res'], then you need to do something like this - data=JSON.parse(data)['res']; now you can get id from data[0]

answered Jul 19, 2018 at 14:46

Ashitaka

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Comments

atta afridi · Accepted Answer · 2018-07-19 13:57:42Z

0

I think the query return empty please try this Code.....

 public function get_menu_check($root_id)
        {
                $data = $this->db->select('C1.menu_code,C1.menu_name')
                                 ->from('create_menu as C1')
                                 ->where('C1.menu_root_id',$root_id)
                                 ->order_by('C1.menu_code')
                                 ->get();
               if($data->num_rows() >= 0)
                  return $data->result_array();
               else
                  return false;
        }

answered Jul 19, 2018 at 13:57

atta afridi

965 bronze badges

5 Comments

sunshine Over a year ago

still it displays undefined

atta afridi Over a year ago

please use last_query function.to check the query result like

atta afridi Over a year ago

$this->db->last_query(); add this statement after get(); function and copy the result and direct check in phpmyadmin.if the query give you some result then the query is correct and the error was in your controller.

zarif khan Over a year ago

on ajax call in success just put console.log(data) at the top of function and watch the console. what is return ?

sunshine Over a year ago

got response in console but how to display in view file? {menu_code: "2", menu_name: "Plant"} {menu_code: "3", menu_name: "Line"}

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