[英]model in codeigniter won't recognize id number
所以這是錯誤:
A PHP Error was encountered
Severity: Notice
Message: Undefined variable: idnum
Filename: models/model_teacher.php
Line Number: 8
這是我的控制器:
public function teacher(){
$this->load->model('model_teacher');
$data['result'] = $this->model_teacher->scoreboard();
$this->load->view('teacher/teacher', $data);
}
和我的模型:
class Model_teacher extends CI_Model {
public function scoreboard() {
$this->db->where('login_id', $this->input->post('idnum'));
$query = $this->db->query("SELECT * FROM teacher WHERE login_id = '".$idnum."'");
return $query->result();
}
}
和視圖:
<?php
foreach ($result as $row) {
echo $row->login_id."<br>";
echo $row->lname."<br>";
}
?>
而且不知道這段代碼有什么問題。 請多多包涵,我仍然是使用Codeigniter的新手。 多謝你們。
更改如下:
在控制器中
public function teacher(){
$this->load->model('model_teacher');
$idNum = $this->input->post('idnum');
$data['result'] = $this->model_teacher->scoreboard($idNum);
$this->load->view('teacher/teacher', $data);
}
在模型中
public function scoreboard($idNum) {
$this->db->where('login_id', $idNum);
$query = $this->db->query("SELECT * FROM teacher WHERE login_id = '".$idnum."'");
return $query->result();
}
如果將它從控制器傳遞到模型會更好
$id = $this->input->post('idnum');
$data['result'] = $this->model_teacher->scoreboard($id);
和型號:
public function scoreboard($id) {
//$this->db->where('login_id', $id);
$query = $this->db->query("SELECT * FROM teacher WHERE login_id = '".$id."'");
return $query->result();
}
另外,您還沒有在$idnum
定義任何位置,因此出現通知
您不能直接進入模型,而必須首先進入控制器
public function teacher(){
$id = $this->input->post('idnum');
$this->load->model('model_teacher');
$data['result'] = $this->model_teacher->scoreboard($id);
$this->load->view('teacher/teacher', $data);
}
模型 `
public function scoreboard($idnum) {
$query = $this->db->query("SELECT * FROM teacher WHERE login_id = '".$idnum."'");
return $query->result();
}`
這是您可以使用的代碼
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