我动态构建HTML页面,除了Twitter共享按钮多次显示外,一切都很顺利。
我查看了网页来源,并没有向我跳出来。
这里是吐出Facebook和Twitter按钮的代码。作为一个数据库人员,我很少知道它意味着什么,因为它只是这些供应商提供的示例代码的剪切和粘贴。
<p>
<div id="fb-root"></div>
<script>(function(d, s, id)
{ var js, fjs = d.getElementsByTagName(s)[0];
if (d.getElementById(id)) return;
js = d.createElement(s);
js.id = id; js.src = "//connect.facebook.net/en_US/all.js#xfbml=1&appId=620849677946889";
fjs.parentNode.insertBefore(js, fjs);}
(document, 'script', 'facebook-jssdk'));
</script>
<div class="fb-like" data-href="https://www.yahoo.com" data-width="400" data-show-faces="True" data-send="True" data-colorscheme="light" data-layout="standard" data-action="like"></div></p>
<p>
<a href="https://twitter.com/share" class="twitter-share-button"data-related="whitehouse,Zerohedge" data-lang="en" data-size="large" data-hashtags="lawyers,guns,money" data-count="horizontal" data-text="Stuff" data-URL="Stuff" data-via="http://www.zerohedge.com"</a>
<script>
!function(d,s,id){
var js,fjs=d.getElementsByTagName(s)[0],p=/^http:/.test(d.location)?'http':'https';
if(!d.getElementById(id)){
js=d.createElement(s);
js.id=id;js.src=p+'://platform.twitter.com/widgets.js';
fjs.parentNode.insertBefore(js,fjs);}}
(document, 'script', 'twitter-wjs');</script></p>
有什么想法吗?
答案 0 :(得分:0)
我不确定你为什么这么多。 我重新生成它,它现在似乎有效。
以下是代码:
<p><div id="fb-root"></div><script>(function(d, s, id) { var js, fjs = d.getElementsByTagName(s)[0]; if (d.getElementById(id)) return; js = d.createElement(s); js.id = id; js.src = "//connect.facebook.net/en_US/all.js#xfbml=1&appId=620849677946889"; fjs.parentNode.insertBefore(js, fjs);}(document, 'script', 'facebook-jssdk'));</script><div class="fb-like" data-href="https://www.yahoo.com" data-width="400" data-show-faces="True" data-send="True" data-colorscheme="light" data-layout="standard" data-action="like"></div></p>
<a href="https://twitter.com/share" class="twitter-share-button" data-url="http://www.zerohedge.com" data-via="James0888" data-size="large">Tweet</a><script>!function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0],p=/^http:/.test(d.location)?'http':'https';if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src=p+'://platform.twitter.com/widgets.js';fjs.parentNode.insertBefore(js,fjs);}}(document, 'script', 'twitter-wjs');</script>
您可能只想在Twitter的网站上重新生成它。
答案 1 :(得分:0)
你<a>
标签不合适
替换
<a href="https://twitter.com/share" class="twitter-share-button" data-related="whitehouse,Zerohedge" data-lang="en" data-size="large" data-hashtags="lawyers,guns,money" data-count="horizontal" data-text="Stuff" data-URL="Stuff" data-via="http://www.zerohedge.com"</a>
带
<a href="https://twitter.com/share" class="twitter-share-button" data-related="whitehouse,Zerohedge" data-lang="en" data-size="large" data-hashtags="lawyers,guns,money" data-count="horizontal" data-text="Stuff" data-URL="Stuff" data-via="http://www.zerohedge.com">some text??</a>