Question

我是新手，出于安全目的尝试使用准备好的语句时遇到错误。致命错误：未捕获错误：调用成员函数prepare（）中的null

我正在尝试使用从HTML表单捕获的数据来设置一个“简单的”准备好的语句来更新数据库记录。 InnoDB v10。我认为我的变量和/或其他语法可能存在问题。任何帮助，我们都感激不尽。：-）

<!DOCTYPE html>
<html>
<body>

<?php   



$hostname = "***";
$username = "***";
$password = "***";
$db = "***";

$dbconnect=mysqli_connect($hostname,$username,$password,$db);

if ($dbconnect->connect_error) {
  die("Database connection failed: " . $dbconnect->connect_error);
}

if(isset($_POST['submit'])) {
  $RedeemableCode=$_POST['RedeemableCode'];
  $FirstName=$_POST['FirstName'];
  $LastName=$_POST['LastName'];
  $Address1=$_POST['Address1'];
  $Address2=$_POST['Address2'];
  $City=$_POST['City'];
  $State=$_POST['State'];
  $Zip=$_POST['Zip'];
  $Country=$_POST['Country'];



$stmt = $mysqli->prepare("UPDATE Orders SET RedeemableCode = ?,FirstName = ?,LastName = ?,Address1 = ?,Address2 = ?,City = ?,State = ?,Zip = ?,Country = ?,OrderDate = NOW() WHERE Status='Not Redeemed' and RedeemableCode = '$RedeemableCode' ");

$stmt->bind_param("sssssssss", $_POST['RedeemableCode'],         $_POST['FirstName'] , $_POST['LastName'], $_POST['Address1'], $_POST['Address2'], $_POST['City'], $_POST['State'], $_POST['Zip'], $_POST['Country']);

$ stmt-> execute（）;

$affected_rows= mysqli_stmt_affected_rows ($stmt);

  if ($affected_rows ==1){
    echo 'Thanks for your order.  Your order is scheduled to arrive in  4-6 weeks.';
        mysqli_stmt_close($stmt);
    mysqli_close($dbconnect);
   }  else {
      echo 'Ooops.  The code that you have entered is either not valid  or was previously redeemed.';
      echo mysqli_error();

      mysqli_stmt_close($stmt);
      mysqli_close($dbconnect);
   }






}

?>




</body>
</html>

MySQL准备语句的问题

0 个答案: