下面是我的代码:
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();
我在最后一行得到的错误是:调用未定义的方法mysqli_stmt::get_result()
以下是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();
}
}
}
如果我写下这一行:
if(!stmt) echo 'Statement prepared'; else echo 'Statement NOT prepared';
它输出未准备的语句‘’。如果我直接在IDE中运行查询,替换?带有值的标记,它工作得很好。请注意,$conn对象在项目中的其他查询中运行良好。
有什么需要帮忙的......
https://stackoverflow.com/questions/8321096
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