Question

我想从数据库表中获取用户（当前登录）的电子邮件和他/她的uid，然后将其插入另一个表中。我已经尝试了但是我得到的空白结果是在uid和电子邮件中。

 <?php

    session_start();


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

     $bidamount = $_POST['bidamount'];
     $email = $_SESSION['$u_email'];
     $uid = $_SESSION['$u_uid'];

     //TO ALERT SUBMISSION OF BLANK FIELDS(IT DOESN'T PREVENT SUBMISSION OF BLANK FIELD THOUGH)
     if (!$bidamount){
         echo "can't submit blank fields";
     }

     //TO CONFIRM YOU ARE CONNECTED TO YOUR DATABASE (OPTIONAL)
     $connection = mysqli_connect('localhost', 'root', '', 'tickmill_auctions');
     if ($connection){
         header ("Location: ../Afterlogin.php?action=success");
     }else{
         die("connection failed");
     }
     //TO INSERT username and password from field to jossyusers database
     $query = "INSERT INTO orders(bidamount,email,uid) VALUES('$bidamount','$email','$uid')";
     $result = mysqli_query($connection, $query);
     if(!$result){
         die("OOPPS! query failed".mysqli_error($connection));
      }
    }

    ?>

Answer 1

这看起来不对，因为我在脚本中看不到名称为$u_email和$u_uid

的变量

$email = $_SESSION['$u_email'];
$uid = $_SESSION['$u_uid'];

删除$标志并尝试

 $email = $_SESSION['u_email'];
 $uid = $_SESSION['u_uid'];

或者，如果你确实设置了这些变量，则使用双引号而不是像这样的单引号

$email = $_SESSION["$u_email"];
$uid = $_SESSION["$u_uid"];

警告：您的脚本对SQL Injection Attack开放甚至if you are escaping inputs, its not safe! 使用prepared parameterized statements

同时测试时，添加

    ini_set('display_errors', 1); 
    ini_set('log_errors',1); 
    error_reporting(E_ALL); 
    mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

到你的脚本的顶部。这将强制执行任何mysqli_错误生成您可以在浏览器上看到的异常，并且您的浏览器上也会显示其他错误。

如何获取用户数据并将其插入另一个表

1 个答案: