Question

I am fetching data in array and its get data when I use print_r function but when I display the array its said undefined data field.
any way to display result array();? I don’t want to use a foreach loop.

The code is:

$menu2=$this->load->common_model->getRecord('*','inner_menu','menu_page',2);
if ( $menu2->num_rows() > 0) { 
  $page1    =   $menu2->result_array();
  print_r($page1);
  <?=$page1['url']?>// here it said undeifned variable url
                    // url is in my db tabel inner_menu
}

Answer 1

Your syntax is really off, or you’re using CodeIgniter in a really strange or unacceptable way.

First, you should be loading your models like this:

$this->load->model('common_model');
$this->common_model->getRecord();

Your getRecord() method should be one that throws result_array() to the controller and not the controller calling upon it.

Second, in order to get a better understanding, we would need to see some sample code of getRecord() for us to help you.