加入得到4个表所有结果由id

Problem

Now I am doing edit form in codeigniter and I want to get all four tables data. From controller comes event_id and by this id I want to get data.

Model

public function get_list_for_edit($event_id){
    return

   $this->db->select('*')
            ->from('events e')
            ->join('events_location el','el.events_id =' $event_id,'left')
            ->join('events_photos ep',
                    'ep.events_id =' $event_id,'left')
            ->join('push_notifications pn',
                    'pn.events_id =' $event_id,'left')
            ->where('e.event_id =' $event_id)
            ->row_object();
 }

Error

Message: syntax error, unexpected '$event_id' (T_VARIABLE)

could be you need string concatenation (othewise the variable in not correctly added to the rest of the string)

  public function get_list_for_edit($event_id){
      return

     $this->db->select('*')
              ->from('events e')
              ->join('events_location el','el.events_id =' . $event_id,'left')
              ->join('events_photos ep',
                      'ep.events_id =' . $event_id,'left')
              ->join('push_notifications pn',
                      'pn.events_id ='. $event_id,'left')
              ->where('e.event_id =' . $event_id)
              ->row_object();
   }

for Call to undefined method CI_DB_mysqli_driver::row_object() error
You are not selecting table name ... so you should add the get('TABLE_NAME') method for the table/model you need

     $this->db->select('*')
              ->from('events e')
              ->join('events_location el','el.events_id =' . $event_id,'left')
              ->join('events_photos ep',
                      'ep.events_id =' . $event_id,'left')
              ->join('push_notifications pn',
                      'pn.events_id ='. $event_id,'left')
              ->where('e.event_id =' . $event_id)->get('TABLE_NAME')
              ->row_object();