Есть такой код
<?php
require_once('./assets/config/config.php');
$connect = mysqli_connect($config['bd_host'], $config['bd_user'], $config['bd_password'], $config['bd_name']) or die(mysqli_error($connect));
if (isset($_POST['auth_login']))
{
if (empty($_POST['auth_login']))
{
$auth_error = "Введите логин";
}
elseif (empty($_POST['auth_password']))
{
$auth_error = "Введите пароль";
}
else
{
$login = $_POST['auth_login'];
$password = $_POST['auth_password'];
$query = mysqli_query($connect, "SELECT `id` FROM `user_login` WHERE `nickname` = '$login' AND `password` = '$password'");
$answer = mysqli_fetch_array($query);
if (empty($answer['id']))
{
$auth_error = 'Введенные данные не верны';
}
else
{
$_SESSION['login'] = $login;
$_SESSION['id'] = $answer['id'];
$_SESSION['logged'] = 1;
$id_sessian = $_SESSION['id'];
$query = mysqli_query($connect, "UPDATE `accounts`.`user_login` SET `logged` = '1' WHERE `user_login`.`id` = '$id_sessian'");
$auth_error = 'Вы успешно вошли';
}
}
}
elseif ($id_sessian == '1') {
$logout = '
<button class="btn btn-outline-success authorization" data-toggle="modal" data-target="#sign-in">
<i class="fa fa-sign-in" aria-hidden="true"></i> Выйти
</button>';
}
else
{
$logout = '
<button class="btn btn-outline-success authorization" data-toggle="modal" data-target="#sign-in">
<i class="fa fa-sign-in" aria-hidden="true"></i> Войти
</button>
<button class="btn btn-outline-success registration ml-2" data-toggle="modal" data-target="#sign-up">
<i class="fa fa-user-plus" aria-hidden="true"></i>Регистрация
</button>';
}
$auth_error = isset($auth_error) ? $auth_error : NULL;
?>
При исполнении elseif условии не выводиться переменная
elseif ($id_sessian == '1') {
$logout = '
<button class="btn btn-outline-success authorization" data-toggle="modal" data-target="#sign-in">
<i class="fa fa-sign-in" aria-hidden="true"></i> Выйти
</button>';
}