how to store database query result in a variable in view [duplicate]

Question

I want to store my database query result in my view section in a variable.I am trying but not work.

My View code

$cd=''.base_url().'/video/series/';
$count = count($cd);
for ($i = 0; $i < $count; $i++) 
{
  print'"'.$cd[$i][1].'",';  

}

My Controller code

public function series() {

    $result= $this->video_model->series_list();

    return $result;


}

My Model Code

function series_list()
{

    $string = trim($this->input->get_post('term'));
    $query = $this->db->query("SELECT name FROM `series` WHERE name LIKE '%".$string."%'");


    return $query->row_array();

}

$cd=''.base_url().'/video/series/'; not get any array data ,only get blank data

Can you show us the database query and how you're trying to store the result? — Jay Blanchard
– Jay Blanchard, Commented Aug 28, 2014 at 17:33

Naincy · Accepted Answer · 2014-08-28 18:00:27Z

2

You can get post variable value in controller only. But you trying to get it in model. That's wrong.

Controller:

public function series() {
$string = trim($this->input->get_post('term'));    
$data['result']= $this->video_model->series_list($string);
$this->load->view('folder/filename', $data);  
// in your case i think folder= video and filename = series
// in this way you can pass value from controller to view

}

Model:

function series_list($string = null)
{
    if($string != ''){
       $query = $this->db->query("SELECT name FROM `series` WHERE name LIKE '%".$string."%'");
       return $query->row_array();
    }
    else
       return false;
}

View:

<?php
 var_dump($data);
?>

you can get your resultset in view and can play with it as you want.

edited Aug 28, 2014 at 18:00

answered Aug 28, 2014 at 17:36

Naincy

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5 Comments

k10gaurav Over a year ago

@NaincyGupta you need to keep the result variable inside an array and then pass array in view.

Naincy Over a year ago

@k10gaurav: he is trying to get post variable in model which is wrong.

k10gaurav Over a year ago

hmm... right @NaincyGupta but variable accessibility requires to be in array so that if I have n number of variables in controller they would be accessible in view.

Naincy Over a year ago

@k10gaurav: I have made changes in my code accordingly. :)

k10gaurav Over a year ago

@NaincyGupta me too. Happy coding :)

k10gaurav · Accepted Answer · 2014-08-28 18:07:48Z

You need to pass variable inside some array(in view method) so that all variables inside common array would be accessible on view page

Controller code would be

public function series() 
{
  $inputData     =trim(strip_tags($this->input->get_post('term')));
  $data['result']= $this->video_model->series_list($inputData);
  $this->load->view('directory/viewpage',$data); //$data is array with all variables inside it
}

Model code would be

function series_list($data)
{
 $string = $data;
 $query = $this->db->query("SELECT name FROM `series` WHERE name LIKE '%".$string."%'");
 if( $query->num_rows>0)
 {
  return $query->row_array();
 }
 else
 {
   return false;
 }
}

View code would be

<?php
     if(isset($result)&&($result!=''))
    {
      echo "<pre/>";
      print_r($result); // Or var_dump($result);
    }
?>

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