PHP bind_param錯誤：在數據庫中插入問號（'？'）而不是實際值

Question

我正在創建註冊表單，並且在將數據插入數據庫時​​使用了PHP綁定參數。

$fnameclean = mysqli_real_escape_string($mysqli_conn, $_POST['first_name']); 
$passwordclean = mysqli_real_escape_string($mysqli_conn, hash("sha512", $_POST['password'])); 
$lnameclean = mysqli_real_escape_string($mysqli_conn, $_POST['last_name']); 
$emailclean= mysqli_real_escape_string($mysqli_conn, $_POST['email']); 



$stmt = $mysqli_conn->prepare("INSERT INTO user (firstname, surname, email, password) VALUES ('?', '?', '?', '?')"); 
$stmt->bind_param("ssss", $fnameclean, $lnameclean, $emailclean, $passwordclean); 

$stmt->execute(); 
$stmt->close(); 

當我按下提交按鈕，我可以在我的數據庫看到在田裏問號：名字，姓氏，電子郵件地址和密碼。

然而，當我嘗試沒有它工作完全正常

代碼綁定參數信息添加到數據庫中：

$query1 = "INSERT INTO user (firstname, surname, email, password) VALUES ('$fnameclean', '$lnameclean', '$emailclean', '$passwordclean')";  
$mysqli_conn->query($query1); 

我在做什麼錯在這裏？

Answer 1

VALUES (?, ?, ?, ?) 

沒有'在查詢中，您使用?用於綁定參數一起使用。所以你的查詢應該像

$stmt = $mysqli_conn->prepare("INSERT INTO user (firstname, surname, email, password) VALUES (?, ?, ?, ?)");