如何在此代碼中使用（名稱）的命令以獲得asc命令的結果

Question

如何在此代碼中使用（名稱）的命令來獲取asc命令的結果？

這是一段代碼：

function getRecordByID($id) 
    { 
     $this->db->select($this->tbl_area_menu.'.*,'.$this->tbl_users.'.firstname as created_by,'.$this->tbl_page.'.name as page_name,'.$this->tbl_page.'.slug as page_slug,'.$this->tbl_page.'.template_directory,'.$this->tbl_directory.'.slug as directory_slug'); 

     $this->db->where($this->tbl_area_menu.'.id',$id); 
     $this->db->join($this->tbl_users,$this->tbl_users.'.id = '.$this->tbl_area_menu.'.created_by','left'); 
     $this->db->join($this->tbl_page,$this->tbl_page.'.id = '.$this->tbl_area_menu.'.page_id','left'); 
     $this->db->join($this->tbl_directory,$this->tbl_directory.'.id = '.$this->tbl_page.'.template_directory','left'); 


     $query = $this->db->get($this->tbl_area_menu); 
     //echo $this->db->last_query(); 
     $record = $query->row(); 
     return $record; 
    } 

Answer 1

看起來你正在使用Query Builder。如果是這樣的話，畢竟join加$this->db->order_by('name', 'ASC');