下面的代码工作正常,如果我明确地加我想用echo
的HTML,但我想使代码更干净由包括标记文件,这是在包括脚本的同一目录中,但由于某种原因它赢得不包括。下面是代码:为什么包含在这里不起作用?
function render_social_links() {
$current_post_id=get_the_ID();
$them_uri = get_stylesheet_directory_uri();
$featured_image_url = '';
if (has_post_thumbnail($current_post_id)) {
$featured_image_id = get_post_thumbnail_id($current_post_id);
$featured_image_url = $featured_image_id ? wp_get_attachment_url($featured_image_id) : '';
}
if(is_home() || is_single()){
if(is_home()){
$bitlink = $lnk = get_bloginfo('url');
$bitly = getBitly($bitlink);
$nam = get_bloginfo('name');
}
elseif(is_single($current_post_id)){
$bitlink = $lnk = get_permalink($current_post_id);
$bitly = getBitly($bitlink);
$nam = get_the_title($current_post_id);
}
include(__DIR__ . '/social-links.php');
}
}
这是社会links.php内的标记,该文件是一个孩子的主题和包含在孩子的functions.php
<div id="socialleft">
<ul>
<li>
<img src="'.$them_uri.'/images/social/share-38.png" alt=""/>
</li>
<li>
<a href="http://www.facebook.com/sharer.php?u='.$lnk.'&t='.$nam.'" title="شارك على فيسبوك" target="_blank">
</li>
<img src="'.$them_uri.'/images/social/facebook-38.png" alt="" />
</a>
</li>
<li>
<a href="http://twitter.com/home/?status='.$nam.' : '.$bitly.'" title="غرد" target="_blank">
<img src="'.$them_uri.'/images/social/twitter-38.png" alt="" />
</a>
</li>
<li>
<a href="https://plus.google.com/share?url='.$lnk.'" onclick="javascript:window.open(this.href,
\'\', \'menubar=no,toolbar=no,resizable=yes,scrollbars=yes,height=600,width=600\');return false;" title="شارك على جوجل+" target="_blank">
<img src="'.$them_uri.'/images/social/Google-plus-38.png" alt="" />
</a>
</li>
</ul>
</div>
你能分享“social-links.php”或类似内容的内容吗?它有一些警告或错误? –
如果我直接使用'social-links.php'的内容而不包含它,那么效果很好。但是当包含任何输出时 –