从数据库中的表中获取一个特定字段并在会话中使用它

问题描述：

I'm currently building a website that has a log in form. Depending on which role this user has a navigation bar has to be displayed accordingly.

For Example : Someone who hasn't yet logged in will have a "login" link. If someone has logged in and is not an admin a different nav bar is displayed.

All log in details are stored in a db and this has to be done using sessions since it will send the user to a different page.

Any ideas ?

Code till now :

$query = "SELECT count(*) FROM tbl_users
            WHERE username = '$username' AND password = '$pass'";

$result = mysqli_query($connection, $query)
        or die("Error in query: ". mysqli_error($connection));

$row = mysqli_fetch_row($result);
$counter = $row[0];

if($counter > 0)
{
    session_start();
    $_SESSION['user']= $username;
    $_SESSION['role'] = $row['role'];

我正在构建一个具有登录表单的网站。根据该用户具有导航栏的角色必须相应地显示。 p>

例如：尚未登录的用户将拥有“登录”链接。如果有人登录且不是管理员，则会显示不同的导航栏。 p>

所有登录详细信息都存储在数据库中，这必须使用会话来完成，因为它会将用户发送到不同的页面。 p>

任何想法？ p>

代码到现在为止： p>

  $ query =“SELECT count（*）FROM tbl_users 
 WHERE username ='$ 用户名'AND密码='$ pass'“; 
 
 $ result = mysqli_query（$ connection，$ query）
或die（”查询错误：“。mysqli_error（$ connection））; 
 
 $  row = mysqli_fetch_row（$ result）; 
 $ counter = $ row [0]; 
 
if（$ counter＆gt; 0）
 {
} session_start（）; 
 $ _SESSION ['user'] = $ 用户名; 
 $ _SESSION ['role'] = $ row ['role']; 
  code>  pre> 
  div>

答

a somewhat vague question but, check your role.

if($_SESSION['role']=='admin')
    echo 'I am admin';
elseif($_SESSION['role']=='customer')
    echo 'I am customer';
else
    echo 'No valid role / not logged in';

从数据库中的表中获取一个特定字段并在会话中使用它

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