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我有一個模型,它返回已登錄到網站的人的用戶名到控制器。我試圖將用戶名保存到一個變量,我可以用戶然後插入到另一個表,但我沒有運氣保存數據。以下是我的模型和控制器類。商店模型函數返回到控制器功能
型號:我控制器
function is_loggedin()
{
$session_id = $this->session->userdata('session_id');
$res = $this->db->get_where('logins',array('session_id' => $session_id));
if ($res->num_rows() == 1) {
$row = $res->row_array();
return $row['name'];
}
else {
return false;
}
}
部分:
public function index()
{
$loggedin = $this->authlib->is_loggedin();
if ($loggedin === false)
$this->load->view('login_view',array('errmsg' => ''));
else
{
$this->load->view('postquestion_view',array('username' => $loggedin));
$user = $loggedin['username'];
}
}
public function askquestion()
{
$qtitle = $this->input->post('title');
$qdetails = $this->input->post('details');
$qtags = $this->input->post('tags');
$qcategory = $this->input->post('category');
$quser = $user;
錯誤:
A PHP Error was encountered
Severity: Notice
Message: Undefined variable: user
Filename: controllers/postq.php
Line Number: 47
請大聲讀出您的控制器並重新執行問題。 (如果'$ loggedin'是'FALSE'呢?) – Kyslik
@Kyslik這就是我最初的代碼(見編輯)。在索引函數中 – Jacob