我对PHP和CodeIgniter完全陌生,我试图在视图中创建一个链接,当单击时将从特定类别返回数据列表。
vgs_model.php模型public function pcList(){
$this->db->select('*');
$this->db->from('videogame');
$this->db->where('Format', 'PC');
$query = $this->db->get();
return $query->result();
}
search.php控制器public function __construct()
{
parent::__construct();
$this->load->helper('form');
$this->load->helper('url');
$this->load->model('vgs_model');
}
public function index()
{
$this->load->view('header_view');
$this->load->view('search_view');
$this->load->view('footer_view');
}
public function getPc(){
$search_term = 'PC';
$data['results'] = $this->Vgs_model->pcList();
$this->load->view('search_results', $data);
}
search_view视图
<a href = <?php echo site_url('Hello/getPc'); ?>>View PC Games</a>
我一直得到以下错误
Message: Undefined property: Search::$Vgs_model
Filename: controllers/search.php
Line Number: 40
第40行是这样$data['results'] = $this->Vgs_model->pcList();
我做错了什么?如有任何帮助,不胜感激。
感谢您阅读我的文章
小写:
$data['results'] = $this->vgs_model->pcList();