如何使用php會話訪問數據庫信息？

Question

我是個nOOb。 我知道如何制作簡單的登錄腳本，但我不太明白如何使用$ _SESSION訪問不同頁面的數據庫信息/列。 我正在嘗試根據user_level顯示一個表。

用戶表

id |username |email |password | user_level | user_roles |last_login 

dashboard.php

<?php if  ($_SESSION['login_user'] == user_level["admin"]):?>

<div class="table">
Admin table
</div>

<?php else: ?>

    <div class="table">
Employee table
</div>

的login.php

<?php
ob_start();
include 'includes/header.php';

 if($_SERVER["REQUEST_METHOD"] == "POST") {
      // username and password sent from form 

      $myusername = mysqli_real_escape_string($db,$_POST['username']);
      $mypassword = mysqli_real_escape_string($db,$_POST['password']); 


      $update_query = "UPDATE users SET last_login=now() WHERE username='" .$myusername. "'";
      $update_q = "UPDATE users SET last_update=now() WHERE username='" .$myusername. "'";
      $sql = "SELECT id FROM users WHERE username = '$myusername' and password = '$mypassword'";
      $execute_update = mysqli_query($db,$update_query);
     $execute_u= mysqli_query($db,$update_q);
      $result = mysqli_query($db,$sql);
      $row = mysqli_fetch_array($result,MYSQLI_ASSOC);
      $active = $row['active'];

      $count = mysqli_num_rows($result);

      // If result matched $myusername and $mypassword, table row must be 1 row

      if($count == 1) {

         $_SESSION['login_user'] = $myusername;

         header("location: dashboard.php");
      }else {
         $error = "Your Login Name or Password is invalid";
      }
 }
?>

session.php文件

<?php
include '../includes/config.php';
   $user_check = $_SESSION['login_user'];

   $ses_sql = mysqli_query($db,"SELECT username FROM users WHERE username = '$user_check' ");

   $row = mysqli_fetch_array($ses_sql,MYSQLI_ASSOC);

   $login_session = $row['username'];

   if(!isset($_SESSION['login_user'])){
      header("location:../login.php");
   }
?>

Answer 1

由於$ _SESSION ['login_user']和數據庫中的用戶名相同，因此可以在多個頁面中使用$ _SESSION變量來獲取登錄用戶的user_level。

//checkUserLevel.php
$user_check = $_SESSION['login_user'];
$ses_sql = mysqli_query($db,"SELECT * FROM users WHERE username = '$user_check' ");//This query will only return 1 user if usernames are unique
while($row = mysqli_fetch_assoc($ses_sql)){
$user_level = $row['user_level']; //Get user_level of that user
}

//Use in dashboard.php
include 'includes/checkUserLevel.php';
if($user_level=='admin'){
 //Show Tables 
}else{
 //Do something else
}

如果不讓我知道你應該能夠使用這種方法，我會嘗試幫助