我在调用表时遇到问题,我得到错误,如何调用表并加载或发送到 View ?
这是代码:
if($page == 'index') {
$this->loggedIn();
} else if($page == 'register') {
$this->loggedRegisterIn();
}
else {
$this->notLoggedIn();
$this->load->library('session');
$this->load->model('model_users');
$data['userData'] = $this->model_users->fetchUserData($this->session->userdata('user_id'));
$data2['userData2'] = $this->model_users->fetchUserData($this->session->userdata('user_id'));
}
$this->load->view($page, $data, $data2);
这是错误信息:
A PHP Error was encountered
Severity: Notice
Message: Undefined variable: userData2
Filename: views/dashboard.php
Line Number: 179
Backtrace:
File: C:\xampp\htdocs\cobaan\e-magang\application\views\dashboard.php Line: 179 Function: _error_handler
File: C:\xampp\htdocs\cobaan\e-magang\application\controllers\Pages.php Line: 32 Function: view
File: C:\xampp\htdocs\cobaan\e-magang\index.php Line: 315 Function: require_once
最佳答案
这应该是(错误)
$data['userData'] = $this->model_users->fetchUserData($this->session->userdata('user_id'));
$data2['userData2'] = $this->model_users->fetchUserData($this->session->userdata('user_id'));
}
$this->load->view($page, $data, $data2);
像这样(正确)
$data['userData'] = $this->model_users->fetchUserData($this->session->userdata('user_id'));
$data['userData2'] = $this->model_users->fetchUserData($this->session->userdata('user_id'));
}
$this->load->view($page, $data); # pass only one array with different array pointer
所以在视野中
如果 foreach
foreach($userData as $item)
{
}
foreach($userData2 as $item)
{
}
如果 print_r
print_r($userData);
print_r($userData2);
关于php - 如何从 codeigniter 调用两个或多个表并发送或加载以查看,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42929762/