可能重複:
PHP Error: mysql_fetch_array() expects parameter 1 to be resource, boolean given警告:mysql_fetch_assoc()預計參數1是資源,鑑於對象
我似乎無法找出什麼我'做錯了。所以,當我提出我的形式我得到警告錯誤和
注意:未定義的變量:在/Library/WebServer/Documents/ArturoLuna_Final/loginCheck.php線30
$username = $_POST['username'];
$password = $_POST['password'];
if($username&&$password)
{
require 'conn.php';
$query = "SELECT * FROM users WHERE username='$username'";
$result = $mysql->query($query) or die(mysqli_error($mysql));
$numrows = $result->num_rows;
if ($numrows!=0)
{
while($row = mysql_fetch_assoc($result))
{
$dbusername = $row['username'];
$dbpassword = $row['password'];
}
//check to see if they match!
if($username==$dbusername&&$password==$dbpassword)
{
echo "youre In!";
}
else
echo "incorrect password!";
}
else
die("that user is dead");
//echo $numrows;
}
else
echo ("Please Enter Username")
數據庫用戶名是什麼我可能會做錯嗎?
又是哪一行是第30行? – ajreal 2011-12-15 14:24:40
請注意:在將用戶發送到數據庫之前,您不要轉義用戶名......閱讀「SQL注入」。 – EGOrecords 2011-12-15 14:28:44
你應該看看使用準備好的語句 – Frank 2011-12-15 15:37:42