在此上下文中，元素li不允许作为元素div的子元素。(抑制此子树中的进一步错误。)

Question

我目前正在开发一个网站，..When使用validator.w3验证站点，它有这样的错误：

Element li not allowed as child of element div in this context. (Suppressing further errors from this subtree.)
…widget-2" class="widget Social_Widget"><div class="socialmedia-buttons smw_lef…
Contexts in which element li may be used:
Inside ol elements.
Inside ul elements.
Inside menu elements whose type attribute is in the toolbar state.
Content model for element div:
Flow content.

我在it..but上找到了一个安装在社交媒体小部件上的插件--我找不到它的具体位置，我找到了社会化-widget.php，但仍然找不到这一行：

<li id="social-widget-2" class="widget Social_Widget"><div class="socialmedia-buttons smw_left"><a href="https://www.facebook.com/JustSimplyOutsourcingWorldwideInc" rel="nofollow" target="_blank"><img width="32" height="32" src="http://www.justsimplyoutsourcingworldwide.com/wp-content/plugins/social-media-widget/images/default/32/facebook.png"
alt=" Facebook"
title=" Facebook" style="opacity: 0.8; -moz-opacity: 0.8;" class="fade" /></a><a href="https://plus.google.com/b/100887468338831289783/100887468338831289783/posts" rel="publisher" target="_blank"><img width="32" height="32" src="http://www.justsimplyoutsourcingworldwide.com/wp-content/plugins/social-media-widget/images/default/32/googleplus.png"
alt=" Google+"
title=" Google+" style="opacity: 0.8; -moz-opacity: 0.8;" class="fade" /></a><a href="https://twitter.com/JSOWorldwide" rel="nofollow" target="_blank"><img width="32" height="32" src="http://www.justsimplyoutsourcingworldwide.com/wp-content/plugins/social-media-widget/images/default/32/twitter.png"
alt=" Twitter"
title=" Twitter" style="opacity: 0.8; -moz-opacity: 0.8;" class="fade" /></a><a href="http://www.linkedin.com/company/just-simply-outsourcing" rel="nofollow" target="_blank"><img width="32" height="32" src="http://www.justsimplyoutsourcingworldwide.com/wp-content/plugins/social-media-widget/images/default/32/linkedin.png"
alt=" LinkedIn"
title=" LinkedIn" style="opacity: 0.8; -moz-opacity: 0.8;" class="fade" /></a><a href="http://instagram.com/jsoworldwide/" rel="nofollow" target="_blank"><img width="32" height="32" src="http://www.justsimplyoutsourcingworldwide.com/wp-content/plugins/social-media-widget/images/default/32/instagram.png"
alt=" Instagram"
title=" Instagram" style="opacity: 0.8; -moz-opacity: 0.8;" class="fade" /></a><a href="http://pinterest.com/jsoworldwide/" rel="nofollow" target="_blank"><img width="32" height="32" src="http://www.justsimplyoutsourcingworldwide.com/wp-content/plugins/social-media-widget/images/default/32/pinterest.png"
alt=" Pinterest"
title=" Pinterest" style="opacity: 0.8; -moz-opacity: 0.8;" class="fade" /></a><a href="http://www.youtube.com/user/JSOutsourcingWw" rel="nofollow" target="_blank"><img width="32" height="32" src="http://www.justsimplyoutsourcingworldwide.com/wp-content/plugins/social-media-widget/images/default/32/youtube.png"
alt=" YouTube"
title=" YouTube" style="opacity: 0.8; -moz-opacity: 0.8;" class="fade" /></a></div></li>

这段代码是通过检查元素看到的。我需要找到那一行，但我在社交网站-widget.php中找不到。

Answer 1

根据错误消息，我猜您在div标记中使用了li标记，这是错误的，请仔细检查代码。

你试着做这个

   <div>
        <li></li>
        <li></li>
        <li></li>
   </div>

ol li必须在 ul 或ul标记中。

   <div>
        <ol>
            <li></li>
            <li></li>
            <li></li>
        </ol>
   </div>