我的英语不好,对任何错误都很抱歉。我是Codeigniter的新手,我正在尝试使用此代码搜索记录,这是我的控制器代码:
public function search() {
if(isset($_POST['search']))
{
$s = $_POST['search'];
$this->db->select('u_id,name,email,phone');
$this->db->from('user');
$this->db->where('name =' ,"{$s}" );
$query = $this->db->get();
$res = $query->result();
$data['user'] = null ;
if($res)
{
$data['user'] = $res ;
}
$this->load->view('filter' , $data);
}
}
它工作正常,但我想在单独的模型中编写此代码。
$this->db->select('u_id,name,email,phone');
$this->db->from('user');
$this->db->where('name =' ,"{$s}" );
$query = $this->db->get();
$res = $query->result();
但我不知道如何将此变量值$s = $_POST['search'];
传递给我的模型。有什么建议吗?
答案 0 :(得分:0)
您可以针对您的问题尝试此解决方案:
public function search() {
$search_value = $this->input->post('search'));
$this->db->select('u_id,name,email,phone');
$this->db->from('user');
if (isset($search_value) && !empty($search_value)) {
$this->db->where('name',$search_value);
// OR
$this->db->like('name', $search_value);
}
$query = $this->db->get();
$res = $query->result();
$data['user'] = null;
if ($res) {
$data['user'] = $res;
}
$this->load->view('filter', $data);
}
我希望它会有所帮助。
答案 1 :(得分:0)
public function search() {
$search = $this->input->post('search');
$this->load->model('your_model'); //<-- You can load the model into "__construct"
$search_result = $this->your_model->your_search_function($search); //<-- This is how you can pass a variable to the model from controller
$this->load->view('your_view',['search_result'=>$search_result]); //<-- This is how you can load the data to the view. Hear your search result array is loaded on your view
}
public function your_search_function($search) {
//You will receive this search variable ^^ here in your model
$this->db->select('u_id,name,email,phone');
$this->db->from('user');
$this->db->where('name',$search);
$query = $this->db->get();
$result = $query->result_array();
if(isset($result) && !empty($result))
{
return $result;
} else
{
return FALSE;
}
}