Php 调用成员函数时出错
当我登录时,我试图为codeigniter项目设置一些用户活动,以便查看哪些用户登录和注销 在我的库函数登录时我在下面有这些数据,但现在抛出错误不确定如何修复它Php 调用成员函数时出错,php,codeigniter,Php,Codeigniter,当我登录时,我试图为codeigniter项目设置一些用户活动,以便查看哪些用户登录和注销 在我的库函数登录时我在下面有这些数据,但现在抛出错误不确定如何修复它 $activity_data = array( 'user_id' => $this->CI->session->userdata('user_id'), 'name' => $this->CI->session->userdata('firstname') . ' ' . $this-&
$activity_data = array(
'user_id' => $this->CI->session->userdata('user_id'),
'name' => $this->CI->session->userdata('firstname') . ' ' . $this->CI->session->userdata('lastname')
);
$this->CI->load->model('admin/common/user_login_model');
// Error Here Line 27
$this->user_login_model->addActivity('login', $activity_data);
错误
在型号上
public function addActivity($key, $data) {
$data['user_id'] = $this->user->getID();
if (isset($data['user_id'])) {
$user_id = $data['user_id'];
} else {
$user_id = 0;
}
$this->db->query("INSERT INTO `" . $this->db->dbprefix . "user_activity` SET
`user_id` = '" . (int)$user_id . "',
`key` = " . $this->db->escape($key) . ",
`data` = " . $this->db->escape(serialize($data)) . ",
`ip` = " . $this->db->escape($this->input->server('REMOTE_ADDR')) . ",
`date_added` = NOW()
");
}您似乎创建了一个CI实例并将模型加载到其中。但是,您尝试使用
$this->user\u login\u model
直接从库类访问模型,这是错误的。请尝试:
$this->CI->user_login_model->addActivity('login', $activity_data);
$this->CI->user_login_model->addActivity('login', $activity_data);