菲爾雜貨店CRUD外科醫生模板

Question

HII IHAVE一個問題如何使用食品污物與菲爾斯模板庫，這裏是我的代碼控制器

public function user_management() { 
    $crud = new grocery_CRUD(); 
    $crud->set_subject('User'); 
    $output = $crud->render(); 
    $this->template->set_layout('blog'); 
    $this->template->set('output', $output); 
    $this->template->title('home', 'Grocery-Crud'); 
    $this->template->build('grocery'); 
} 

這裏是我的看法

<?php foreach($css_files as $file): ?> 
<?php endforeach; ?> 
<?php foreach($js_files as $file): ?> 
<?php endforeach; ?> 
<?php echo $output; ?> 

我得到這樣的錯誤

A PHP Error was encountered Severity: Notice Message: Undefined variable: css_files Filename: views/grocery.php Line Number: 2 A PHP Error was encountered Severity: Warning Message: Invalid argument supplied for foreach() Filename: views/grocery.php Line Number: 2 A PHP Error was encountered Severity: Notice Message: Undefined variable: js_files Filename: views/grocery.php Line Number: 5 A PHP Error was encountered Severity: Warning Message: Invalid argument supplied for foreach() Filename: views/grocery.php Line Number: 5 A PHP Error was encountered Severity: 4096 Message: Object of class stdClass could not be converted to string Filename: views/grocery.php Line Number: 10 

請幫助

Answer 1

控制器

public function user_management() { 
$crud = new grocery_CRUD(); 
$crud->set_table('user'); 
$crud->set_subject('User'); 
$output = $crud->render(); 
$this->template->set_layout('blog'); 
$this->template->set('output', (array) $output); 
$this->template->title('home', 'Grocery-Crud'); 
$this->template->build('grocery'); 
} 

鑑於

<?php 
foreach($output['css_files'] as $file): ?> 
<link type="text/css" rel="stylesheet" href="<?php echo $file; ?>" /> 
<?php endforeach; ?> 
<?php foreach($output['js_files'] as $file): ?> 
<script src="<?php echo $file; ?>"></script> 
<?php endforeach; ?> 

<div> 
    <?php echo $output['output']; ?> 
</div>