阻止在当前页面上触发函数

我创建了一个帖子列表，每个帖子都是关于一个车型的。之后，我写了一篇关于最佳车型的帖子。这篇文章抓住了关于每一款车型的每一篇文章，并将它们以片段列表格式显示在主文章“最佳车型”中。

我正在使用下面的功能显示社交共享按钮。我想从显示社交按钮中排除当前帖子，以便按钮仅显示在帖子片段上。

例如获取当前帖子id并使用if语句对其进行过滤。谢谢

`功能crunchify\\u social\\u sharing\\u按钮（$内容）{

        // Get current page URL 
        $crunchifyURL = urlencode(get_permalink());

        // Get current page title
        $crunchifyTitle = htmlspecialchars(urlencode(html_entity_decode(get_the_title(), ENT_COMPAT, \'UTF-8\')), ENT_COMPAT, \'UTF-8\');
        // $crunchifyTitle = str_replace( \' \', \'%20\', get_the_title());

        // Get Post Thumbnail for pinterest
        $crunchifyThumbnail = wp_get_attachment_image_src( get_post_thumbnail_id( $post->ID ), \'full\' );

        // Construct sharing URL without using any script
        $twitterURL = \'https://twitter.com/intent/tweet?text=\'.$crunchifyTitle.\'&amp;url=\'.$crunchifyURL.\'&amp;via=dica.com.br\';
        $facebookURL = \'https://www.facebook.com/sharer/sharer.php?u=\'.$crunchifyURL;
        $whatsappURL = \'https://wa.me/?text=\'.$crunchifyTitle . \' \' . $crunchifyURL;


        // Based on popular demand added Pinterest too
        $pinterestURL = \'https://pinterest.com/pin/create/button/?url=\'.$crunchifyURL.\'&amp;media=\'.$crunchifyThumbnail[0].\'&amp;description=\'.$crunchifyTitle;

        // Add sharing button at the end of page/page content


        $content .= \'<ul class="ul">\';
        $content .= \'<li class="twitter"> <a target="_blank" rel="noopener nofollow" href="\'. $twitterURL .\'" target="_blank"><svg class="icon text-btn-icon svg-inline--fa fa-w-18 fa-fw" style="fill: #fff; "><use xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="#icon-twitter"></use></svg> <span>Twitter </span></a>\';
        $content .= \'<li class="facebook"> <a target="_blank" rel="noopener nofollow" href="\'.$facebookURL.\'" target="_blank"><svg class="icon text-btn-icon svg-inline--fa fa-w-18 fa-fw" style="fill: #fff; "><use xmlns:xlink="http://www.w3.org/1999/xlink" xlink:href="#icon-facebook-share"></use></svg> <span>Facebook </span></a>\';
        $content .= \'<li id="hidewhatsapp" class="whatsapp"> <a href="\'.$whatsappURL.\'" target="_blank"><svg class="icon text-btn-icon svg-inline--fa fa-w-18 fa-fw" style="fill: #fff;  viewBox="0 0 100 100" class="share-icon"> <use xlink:href="#icon-whatsapp"></use></svg> <span> WhatsApp </span></a>\';




        $content .= \'</ul> </div>\';


        return $content;



}
add_shortcode( \'share-buttons\', \'crunchify_social_sharing_buttons\' );`

Get category url in loop

我需要获取类别url，以便将其放入元代码段中。现在，span内的输出是带有url的标记。我只需要获取不带标记的url，并将其放入$cat\\u display中。我尝试使用2个选项，但会看到一个错误：注意：尝试获取非对象的属性// Get post category info $category = get_the_category(); if(!empty($category)) { // Get last category post is in

相关推荐

Get category url in loop