这是我的代码:
include 'conn.php';
$conn = new Connection();
$query = 'SELECT EmailVerified, Blocked FROM users WHERE Email = ? AND SLA = ? AND `Password` = ?';
$stmt = $conn->mysqli->prepare($query);
$stmt->bind_param('sss', $_POST['EmailID'], $_POST['SLA'], $_POST['Password']);
$stmt->execute();
$result = $stmt->get_result();
我在最后一行收到以下错误:调用未定义的方法mysqli_stmt :: get_result()
这是conn.php的代码:
define('SERVER', 'localhost');
define('USER', 'root');
define('PASS', 'xxxx');
define('DB', 'xxxx');
class Connection{
/**
* @var Resource
*/
var $mysqli = null;
function __construct(){
try{
if(!$this->mysqli){
$this->mysqli = new MySQLi(SERVER, USER, PASS, DB);
if(!$this->mysqli)
throw new Exception('Could not create connection using MySQLi', 'NO_CONNECTION');
}
}
catch(Exception $ex){
echo "ERROR: ".$e->getMessage();
}
}
}
如果我写这行:
if(!stmt) echo 'Statement prepared'; else echo 'Statement NOT prepared';
打印“未准备的声明”。如果我直接在IDE中运行查询替换?用值标记,效果很好。请注意,$ conn对象在项目中的其他查询中可以正常工作。
请帮忙......
$_POST['EmailID'], $_POST['SLA'], $_POST['Password']
提交正确使用与POST方法的HTML表单
$stmt = $conn->mysqli->stmt_init();
吗?