由於某種原因,我在視圖中獲取了一個未定義的變量,並不確定原因。當我做一個footer_links變量的print_r時,它顯示正常:未定義變量footer_links
Array([0] => stdClass Object([link_name] => Home [short_name] => index)1 => stdClass Object [link_name] =>關於我們[short_name] => about-us)[2] => stdClass Object([link_name] => Site Map [short_name] => site-map)[3] => stdClass Object([link_name ] =>聯繫我們[=]聯繫我們)[4] => stdClass對象([link_name] =>新聞饋送[短名] =>新聞饋送))
我知道我不是傳遞變量,但當我嘗試與構建它沒有工作。不知道爲什麼我應該把它傳入。我正在使用Phil Sturgeon的模板庫。
A PHP Error was encountered
Severity: Notice
Message: Undefined variable: footer_links
Filename: v1/footer.php
Line Number: 4
A PHP Error was encountered
Severity: Warning
Message: Invalid argument supplied for foreach()
Filename: v1/footer.php
Line Number: 4
控制器:
$activeTemplate = $this->kow->getTemplate();
$siteInfo = $this->kow->getSiteTitleAndSlogan();
$footer_links = $this->kow->getFooterNav();
$this->template
->title($siteInfo[0]->site_name,$siteInfo[0]->site_slogan)
->prepend_metadata('<link rel="stylesheet" type="text/css" href="http://www.kansasoutlawwrestling.com/assets/css/'.$activeTemplate[0]->short_name.'.css" />')
->set_partial('header', $activeTemplate[0]->short_name.'/header')
->set_partial('footer', $activeTemplate[0]->short_name.'/footer')
->build('kow');
查看:
<div id="footer">
<ul>
<?php foreach($footer_links as $row)
{
?>
<li><a><?php echo $row->link_name; ?></a></li>
<?php
}
?>
</ul>
<p>©COPYRIGHT 2010 ALL RIGHTS RESERVED</p>
</div>
編輯:
我發現這個計算器的問題是同一類交易我與...合作。 Creating Dynamic Links Through DB and CI
控制器:
$activeTemplate = $this->kow->getTemplate();
$siteInfo = $this->kow->getSiteTitleAndSlogan();
$footer_links['rows'] = $this->kow->getFooterNav();
$this->template
->title($siteInfo[0]->site_name,$siteInfo[0]->site_slogan)
->prepend_metadata('<link rel="stylesheet" type="text/css" href="http://www.kansasoutlawwrestling.com/assets/css/'.$activeTemplate[0]->short_name.'.css" />')
->set('footer', $footer_links)
->set_partial('header', $activeTemplate[0]->short_name.'/header')
->set_partial('footer', $activeTemplate[0]->short_name.'/footer')
->build('kow');
型號:
function getFooterNav()
{
$this->db->select('site_menu_structures_links.link_name,site_menu_structures_links.short_name');
$this->db->from('site_menu_structures_links');
$this->db->join('site_menu_structures', 'site_menu_structures.id = site_menu_structures_links.menu_structure_id');
$this->db->where('site_menu_structures.short_name', 'footernav');
$query = $this->db->get();
return $query->result_array();
}
查看:
<div id="footer">
<ul>
<?php foreach($rows as $row)
{
?>
<li><a><?php echo $row->link_name; ?></a></li>
<?php
}
?>
</ul>
<p>©COPYRIGHT 2010 ALL RIGHTS RESERVED</p>
</div>
我更新了我的帖子。 –
仔細看看回答該問題的人的第一個代碼塊。我相信這就是你做錯了什麼。它現在工作嗎?請現在發佈錯誤(如果有的話),現在你已經改變了它。 – Spyros