PHP如何根据数据库存储的enum字段值实现登录用户不同页面跳转

• 核心错误：执行SELECT用户查询得到结果集$results后，没有调用mysqli_fetch_assoc()（或同类结果解析函数）将结果集转为数组，代码中用到的$row属于未定义变量，$row['speciality']永远为空，因此判断条件永远不成立，始终走else分支跳转至pro.php。
• 次要错误：$_SESSION["user_name"] = $firstName中的$firstName也未赋值，属于未定义变量，存入session的用户名称会为空。
• 优化建议：跳转后建议新增exit()终止后续代码执行，避免逻辑异常；当前使用md5加密密码安全性极低，且直接拼接SQL语句存在SQL注入风险，后续可改用预处理查询+password_hash/password_verify做密码校验。

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);
include_once("config.php");
include_once("session.php");

if (isset($_POST['signin'])) {
  $email = mysqli_real_escape_string($conn, $_POST['email']);
  $password = mysqli_real_escape_string($conn, $_POST['password']);
  $errors = array();

  if (empty($email)) {
    array_push($errors, "email is required");
  }
  if (empty($password)) {
    array_push($errors, "Password is required");
  }

  if (count($errors) == 0) {
    $password = md5($password);
    $query = "SELECT * FROM user WHERE email='$email' AND password='$password'";
    $results = mysqli_query($conn, $query);
    if (mysqli_num_rows($results) == 1) 
    {
      // 新增代码：解析查询结果为关联数组
      $row = mysqli_fetch_assoc($results);
      // 新增代码：从结果中取firstName赋值
      $firstName = $row['firstName'];

      $_SESSION['email'] = $email;
      $_SESSION["user_name"] = $firstName;
      $_SESSION['success'] = "You are now logged in";
      $logintime = "UPDATE user SET lastLogin = now() where email = '$email'";
      mysqli_query($conn, $logintime);
      
      //Check speciality and redirect accordingly
      $speciality = $row['speciality'];
      if($speciality == "stu"){
        header("location:stu.php");
        // 新增代码：跳转后终止执行
        exit();
      }else{
        header("location:pro.php");
        exit();
      }
    }else 
    {
        array_push($errors, "Wrong username/password combination");
        header("location:login.php");
        exit();
    }
  }
}
?>

内容的提问来源于stack exchange，提问作者pos3id0nas