-4
我需要幫助找出原因,這是錯誤警告:mysqli_query()預計參數1是mysqli的,在C空給出: XAMPP htdocs中決賽上線14
<?php
class data
{
private $db;
public $nama, $password, $alamat, $jk, $kodepos, $alasan, $email;
function _construct($db)
{
$this->db = $db;
}
public function input_data()
{
$query = "INSERT INTO data (nama, password, alamat, jeniskelamin, kodepos, alasan, email)VALUES('$this->nama', '$this->password',
'$this->alamat', '$this->jk', '$this->kodepos', '$this->alasan', '$this->email')";
$insert = mysqli_query($this->db, $query);
return $insert;
}
public function lihat_data()
{
$query = "SELECT * FROM data ORDER BY id";
$view = mysqli_query($this->db, $query);
return $view;
}
}
?>
這意味着連接失敗! – Rizier123
你的問題是什麼? –
你想對這篇文章說些什麼?請改善你的問題! – starkeen