Problem displaying data in view file codeigniter

Question

I have a problem with Codeigniter. I am unable to display data in view file. Given below is the code of my model, controller and view file. I have also attached a screenshot of the error for better clarity.

Model

public function get_projects ($userid = '')
 {
   if (empty($userid)){
     return FALSE;
   }
        $this->db->select('*');
        $this->db->from('projects');
        $this->db->where('userid', $userid);
        $query = $this->db->get();

   return $query->row_array();                 
 }

Controller

$uid = $this->session->userdata('user_id');

        $data['details']= $this->projects->get_projects($uid);
        $this->load->view('design/header', $data);
        $this->load->view('design/nav', $data);
        $this->load->view('content/dashboard', $data);
        $this->load->view('design/footer', $data);

View

<div class="col-6">
            <?php foreach ($details as $row){ echo $row->id; } ?>
        </div>

Error

Have you ever tried to check if it is an associative array instead? The return $query->row_array(); Seeps that returns the data as an associative array. — Dimitrios Desyllas
– Dimitrios Desyllas, Commented Jun 20, 2019 at 18:28
And did it work or not? Is has not worked then update your question with your trial ;) . Also update the question showing the result of var_dump($row); furthermore in theese situations you can use a tool named xdebug as well. — Dimitrios Desyllas
– Dimitrios Desyllas, Commented Jun 20, 2019 at 18:40
it did not work. I also tried changing my model file to if($query->num_rows() == 1){ return $query->row(); }else{ return $query->result(); } — Aman Mishra
– Aman Mishra, Commented Jun 20, 2019 at 18:42
On your view just place var_dump($row) on the first line of it from that figure out what type of variable is. (After you debugged it just remove it). About var_dump() look on documentation. — Dimitrios Desyllas
– Dimitrios Desyllas, Commented Jun 20, 2019 at 18:48

InOrderToLive · Accepted Answer · 2019-06-21 03:10:01Z

1

The problem here is the return of model function: return $query->row_array(); that returns only 1 array (or return $query->row(); that also returns only 1 object). While you are iterating it in view.

Not sure about your purpose, but you can fix it by or:

Do not use iteration in view, just use the variable $details->id
Change your return $query->row_array(); to return $query->result();

answered Jun 21, 2019 at 3:10

InOrderToLive

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1 Comment

Aman Mishra Over a year ago

Yeah I found the problem and now understand it. Thanks a lot.

Alex · Accepted Answer · 2019-06-20 22:03:17Z

First off, it sounds like you want an object as you are using object notation $row->id and not $row['id']. This means you need to use row(). This also means since you will have 1 result, you don't need a foreach.

Model:

public function get_project($userid) {
   $q = $this->db->get_where('projects', array('userid', $userid));
   if ($q->num_rows() !== 1) {
       return false;
   }
   return $q->row();                 
 }

Controller:

$uid = $this->session->userdata('user_id');
if (empty($uid)) {
    show_error('user id not set'); // we probably shouldn't have gotten this far
}
$project = $this->projects->get_project($uid);
if ($project === FALSE) {
    show_error("no rows in project table for user with id: $uid");
}
$data['details'] = $project;
$this->load->view('design/header', $data);
$this->load->view('design/nav', $data);
$this->load->view('content/dashboard', $data);
$this->load->view('design/footer', $data);

View:

<p><?php echo $details->id; ?></p>
<p><?php echo $details->someothervar; ?></p>

While this code may not resolve your problem (I am guessing that there are either no projects for a user with that id or that the session uid is incorrectly set) it will help you debug things and it is good practice to always check for num_rows() and that your variables are outputting what you expect they are. I tend to keep all of these checks in the controller.

thanks Alex for your reply. I came to know some important things through this.

Falyoun · Accepted Answer · 2019-06-21 08:01:05Z

0

In your view page change $row->id; to $row['id'] it will generate result

<div class="col-6">
            <?php foreach ($details as $row){ echo $row['id']; } ?>
 </div>

edited Jun 21, 2019 at 8:01

Falyoun

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answered Jun 21, 2019 at 6:13

deepak ms

1

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