如果已使用其他会话登录，则会中断登录流

Question

我正在开发一个代码点火器 3 应用程序，我最近实现了一个会话检查器，如果用户已经登录，它会删除用户会话。现在，我们希望在用户已经使用另一个会话登录时弹出一个模式框。我能够使用按钮弹出一个模态框，但我想将其实现到登录系统的原始流程中。 因为它是登录表单，因此将您直接带到验证登录系统。这是现在的登录表单：

<form action="<?php echo site_url('login/validate_login/user'); ?>" method="post">
<div class="content-box">
<div class="basic-group">
<div class="form-group">
<label for="login-email"><span class="input-field-icon"><i class="fas fa-envelope"></i></span> <?php echo get_phrase('email'); ?>:</label>
<input type="email" class="form-control" name = "email" id="login-email" placeholder="<?php echo get_phrase('email'); ?>" value="" required>
</div>
<div class="form-group">
<label for="login-password"><span class="input-field-icon"><i class="fas fa-lock"></i></span> <?php echo get_phrase('password'); ?>:</label>
<input type="password" class="form-control" name = "password" placeholder="<?php echo get_phrase('password'); ?>" value="" required>
</div>
</div>
</div>
<div class="content-update-box">
<button type="submit" class="btn"><?php echo get_phrase('login'); ?></button>
</div>
<!-- Modal -->
<div class="modal fade" id="login" role="dialog">
<div class="modal-dialog">

<!-- Modal content-->
<div class="modal-content">
<div class="modal-header">                                
<h4 class="modal-title">You are already logged in</h4>
</div>
<div class="modal-body">
<p>You are currently logged in on a different session on the site. Please note that if you continue, the existing session will be terminated. Please change your password if you suspect that your account has been conpromised.</p>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Cancel Login</button>
<button type="submit" class="btn"><?php echo get_phrase('login'); ?></button>
</div>
</div>

</div>
</div>
<div class="forgot-pass text-center">
<span><?php echo get_phrase('or'); ?></span>
<a href="javascript::" onclick="toggoleForm('forgot_password')"><?php echo get_phrase('forgot_password'); ?></a>
</div>
<div class="account-have text-center">
<?php echo get_phrase('do_not_have_an_account'); ?>? <a href="javascript::" onclick="toggoleForm('registration')"><?php echo get_phrase('sign_up'); ?></a>
</div>
</form>

此刻的按钮直接进入此登录功能：

public function validate_login($from = "") {
$email = $this->input->post('email');
$password = $this->input->post('password');
$credential = array('email' => $email, 'password' => sha1($password), 'status' => 1);
// Checking login credential for admin
$query = $this->db->get_where('users', $credential);
if ($query->num_rows() > 0) {
$row = $query->row();
$this->session->set_userdata('user_id', $row->id);
$this->session->set_userdata('role_id', $row->role_id);
$this->session->set_userdata('role', get_user_role('user_role', $row->id));
$this->session->set_userdata('name', $row->first_name.' '.$row->last_name);
$this->delete_session_user_id();
$this->session->set_flashdata('flash_message', get_phrase('welcome').' '.$row->first_name.' '.$row->last_name);
if ($row->role_id == 1) {
$this->session->set_userdata('admin_login', '1');
redirect(site_url('admin/dashboard'), 'refresh');
}else if($row->role_id == 2){
$this->session->set_userdata('user_login', '1');
$this->set_session_user_id();
redirect(site_url('home/my_courses'), 'refresh');
}
}else {
$this->session->set_flashdata('error_message',get_phrase('invalid_login_credentials'));
redirect(site_url('home/login'), 'refresh');
}
}

我创建了这个函数来从电子邮件中提取用户 ID：

public function get_user_id($user_email = "") {
$this->db->select('id');
$this->db->where('email', $user_email);
$user_id=$this->db->get('users');
return $user_id;
}

此函数可以根据提供的电子邮件获取用户 ID。

然后，我使用此函数检查是否存在会话，如果有 0 个结果，则返回false，如果存在具有该用户 ID 的会话，则返回true。因此，如果它为假，他们应该能够登录，并且模式弹出窗口不应该打开，但如果它是真的，它应该打开。

public function user_has_session($user_id=''){
$this->db->where('user_id',$user_id);
$this->db->from('ci_sessions');
$total=$this->db->count_all_results();
if($total<0) 
return false;
else
return true;
}

我认为这是最好的方法，无需重做整个登录流程。也许有人可以建议这是否是最好的方法，或者实际上我是否应该改变整个流程。

谢谢

这是我之前遇到的问题，我已经回答了自己 https://stackoverflow.com/questions/62458226/codeigniter-3-stop-multiple-logins-using-ci-sessions-database

更新已添加以澄清我的问题

那么问题是当尝试从另一台设备登录时，它应该注销另一个活动会话。因此，如果在新浏览器中登录我的桌面，甚至使用相同的用户ID登录我的手机，则活动会话应该结束，目前它会在没有警告用户的情况下这样做。所以我想有一个模式弹出窗口警告用户当前有一个活动会话正在运行，此用户 ID

Answer 1

您需要实现一个 Ajax 调用，该调用将检查用户是否已登录。如果用户未登录，则可以继续登录，否则会触发弹出窗口打开以显示消息。

在这里，用户可以选择是否登录，如果用户选择登录，则可以在提交时取消绑定事件并让用户继续。

我对您的 HTML 文件进行了一些更改。请查看以下内容 -

您的 HTML 模板

<form action="<?php echo site_url('login/validate_login/user'); ?>" id="login-form" onSubmit="return checkUserSession();" method="post">
<div class="content-box">
<div class="basic-group">
<div class="form-group">
<label for="login-email"><span class="input-field-icon"><i class="fas fa-envelope"></i></span> <?php echo get_phrase('email'); ?>:</label>
<input type="email" class="form-control" name = "email" id="login-email" placeholder="<?php echo get_phrase('email'); ?>" value="" required>
</div>
<div class="form-group">
<label for="login-password"><span class="input-field-icon"><i class="fas fa-lock"></i></span> <?php echo get_phrase('password'); ?>:</label>
<input type="password" class="form-control" name = "password" placeholder="<?php echo get_phrase('password'); ?>" value="" required>
</div>
</div>
</div>
<div class="content-update-box">
<button type="submit" class="btn"><?php echo get_phrase('login'); ?></button>
</div>
<!-- Modal -->
<div class="modal fade" id="login" role="dialog">
<div class="modal-dialog">

<!-- Modal content-->
<div class="modal-content">
<div class="modal-header">                                
<h4 class="modal-title">You are already logged in</h4>
</div>
<div class="modal-body">
<p>You are currently logged in on a different session on the site. Please note that if you continue, the existing session will be terminated. Please change your password if you suspect that your account has been conpromised.</p>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Cancel Login</button>
<button type="button" id="modal-submit-button" class="btn"><?php echo get_phrase('login'); ?></button>
</div>
</div>

</div>
</div>
<div class="forgot-pass text-center">
<span><?php echo get_phrase('or'); ?></span>
<a href="javascript::" onclick="toggoleForm('forgot_password')"><?php echo get_phrase('forgot_password'); ?></a>
</div>
<div class="account-have text-center">
<?php echo get_phrase('do_not_have_an_account'); ?>? <a href="javascript::" onclick="toggoleForm('registration')"><?php echo get_phrase('sign_up'); ?></a>
</div>
</form>

<script>
/** Trigger function on form submit whether to check user logged in */
function checkUserSession(){
var email = $("#login-email").val();
$.ajax({
url: "<?php echo site_url('login/checkUserSession'); ?>",
type: 'POST',
data: { 'email': email},                               
success: function(status){                 
if(status == true) { // User is already logged in somewhere, display the messege.
$("#login").modal();
return false;
} else { // User is not logged in, submit the form
return true;
}
}
});
}

/** Allow user to log in with exception */
$("#modal-submit-button").on("click", function(){
$("#login").modal('hide'); // hide the modal
$("#login-form").attr("onSubmit", ""); // unbind the function
$("#login-form").submit(); // submit login form
})
</script>

控制器-

<?php
/** Function to check user logged in or not */
public function checkUserSession() {
$user_email = $this->input->post('email');
$userId = $this->get_user_id($user_email);
$response = $this->user_has_session($userId);
echo $response;
}
public function get_user_id($user_email = "") {
$this->db->select('id');
$this->db->where('email', $user_email);
$user_id=$this->db->get('users');
return $user_id;
}
public function user_has_session($user_id=''){
$this->db->where('user_id',$user_id);
$this->db->from('ci_sessions');
$total=$this->db->count_all_results();
if($total<0) 
return false;
else
return true;
}
?>