修改退出登录后的提示行为

在FastAdmin中，默认退出登录后会显示"退出成功"的提示信息并跳转页面。要实现不显示提示信息直接跳转，可以通过以下方式修改：

方法一：修改控制器逻辑

找到application/admin/controller/Login.php文件中的logout方法，将原有代码替换为：

/*** 退出登录*/public function logout(){if ($this->request->isPost()) {$this->auth->logout();     Hook::listen("admin_logout_after", $this->request);$this->success(__('Logout successful'), 'index/login');}$html = "<form id='logout_submit' name='logout_submit' action='' method='post'>" . token() . "<input type='submit' value='ok' style='display:none;'></form>";$html .= "<script>document.forms['logout_submit'].submit();</script>";return $html;}

将$this->success(__('Logout successful'), 'index/login');修改为$this->redirect('index/login'); 这样就不会有跳转提示了

 /*** 退出登录*/public function logout(){if ($this->request->isPost()) {$this->auth->logout();     Hook::listen("admin_logout_after", $this->request);$this->redirect('index/login');}$html = "<form id='logout_submit' name='logout_submit' action='' method='post'>" . token() . "<input type='submit' value='ok' style='display:none;'></form>";$html .= "<script>document.forms['logout_submit'].submit();</script>";return $html;}