我有一个 Flash 站点和 php 面板。它的xml文件是动态的。我想在菜单项上添加外部链接。
EG 新闻菜单转到 news.com
我在面板/应用程序/控制器中找到了 process.php。具有以下内容:
//News
if($lang == 'tr')
{
$nTitle = 'Haberler';
$nName = 'Haber Listesi';
}
else
{
$nTitle = 'News';
$nName = 'News List';
}
echo '<page>
<title><![CDATA['.$nTitle.']]></title>
<data>news</data>
'.$this->getpagebackground('news').'
<page>
<title><![CDATA['.$nName.']]></title>
<data>news_list</data>'.$nl;
$newsRs = $this->db->query("Select date_format(date,'%d.%m.%Y') date,content,content_en,id,title,title_en From news Where status = 10 Order By viewOrder Desc");
if($newsRs[0])
{
foreach($newsRs as $news):
if($lang == 'tr')
{
$nContent = $news->content;
$newsTitle = $news->title;
}
else
{
$nContent = $news->content_en;
$newsTitle = $news->title_en;
}
echo '<news>
<data>news_'.$news->id.'</data>
<title>'.$newsTitle.'</title>
<date><![CDATA['.$news->date.']]></date>
<sum>'.substr($nContent,0,150).'</sum>
<content><![CDATA[<p>'.$nContent.'</p>]]></content>'.$nl;
$newsGallery = $this->db->query("Select * From news_media Where status=10 And news_id=".$news->id);
if($newsGallery[0])
{
echo '<gallery>'.$nl;
foreach($newsGallery as $ng):
echo '<item>'.$ng->media.'</item>';
endforeach;
echo '</gallery>'.$nl;
}
echo '</news>'.$nl;
endforeach;
}
echo '</page>'.$nl;
我可以在这里更改链接吗?