這是錯誤說mysqli的錯誤非對象
Fatal error: Call to a member function fetch() on a non-object in C:\xampp\htdocs\Grading\logging.php on line 30
我不知道寫什麼有IM中的mysqli還挺新的,這裏是我的代碼
<?php
$username = $_POST['username'];
$password = $_POST['password'];
$sql = $con->query("SELECT * FROM user WHERE username = '".$username."' ");
$row = $sql->fetch();
if ($username == $row['username'] and $password == $row['password'])
{
echo "<script>window.location.href='home.php'; </script>";
}
else
{
echo "<script> alert ('Invalid Username or Password'); window.location.href='login.php'; </script>";
}
?>
這是整個代碼?您是否在調用函數之前在任何地方創建'$ con'對象? – Oldskool 2013-02-18 08:50:50
$ con = new mysqli('localhost','root','','teacher'); – Goenitz 2013-02-18 09:02:21
更好的使用預製的[mysqli wrapper - safeMysql](https://github.com/colshrapnel/safemysql/blob/master/safemysql.class.php)。它會爲你節省很多頭痛。 – 2013-02-18 09:13:04