Question

我是PHP新手，正在为Web应用程序编写代码。登录和注册功能正常运行，但是我在使用会话ID将数据插入新表时遇到问题，该会话ID在用户注册时自动生成并存储在用户信息表中。这是有问题的功能：

    public function buy($userid, $today_date, $mths, $curr_year, $threefifty, $twofifty, $onefifty, $hundred, $fifty, $t_amnt, $commsn){
     $order = "INSERT INTO bakery_tbl (id, dys, mnths, three_fty, two_fty, one_fty, one_hundred, unit_fifty, amt, comm) VALUES ('$userid', '$today_date', '$mths', '$curr_year', '$threefifty', '$twofifty', '$onefifty', '$hundred', '$fifty', '$t_amnt', '$commsn')";
     $ex_buy = $this->db->query($order) or die(mysqli_connect_error(). "Data cannot be inserted");
 return $ex_buy;
   }

这是实现：

  <?php

   session_start();
   require 'BakeryModel.php';
   $model = new BakeryModel();

  if(isset($_POST['order'])){

 $dy = $_POST['dy'];
 $mnth = $_POST['mnth'];
 $yr = $_POST['yr'];
 $q1 = $_POST['q1'];
 $q2 = $_POST['q2'];
 $q3 = $_POST['q3'];
 $q4 = $_POST['q4'];
 $q5 = $_POST['q5'];


 $a1 = $q1 * 350;
 $a2 = $q2 * 250;
 $a3 = $q3 * 150;
 $a4 = $q4 * 100;
 $a5 = $q5 * 50;

 $total = $a1 + $a2 + $a3 + $a4 + $a5;

 $total_comm = 15*$total/100;
    $purchase = $model->buy($_SESSION['user_id'], $dy, $mnth, $yr, $a1,         $a2, $a3, $a4, $a5, $total, $total_comm);

  if($purchase){
    header("location: appreciate.php");
   }
 } 

 if(isset($_POST['logout'])){   
   $model->log_out();
   header("location: index.php");
  }

?>

<!Doctype html>
//HTML code here.`

每当我尝试提交表单数据时，我都会收到错误号0。 N / B->注销功能正常运行。我还尝试从表单中回显值，它们都显示期望值，包括会话ID。

在插入具有会话ID的数据库时，mysqli错误号为0

0 个答案: