用xslt解析html
有人能帮我拿一下吗用xslt解析html,html,xml,xslt,xslt-1.0,Html,Xml,Xslt,Xslt 1.0,有人能帮我拿一下吗 <rss xmlns:dc="http://purl.org/dc/elements/1.1/" version="2.0"> <channel> <title>This is a test</title> <link>http://somelink.html</link> <description>RSS Feed</description>
<rss xmlns:dc="http://purl.org/dc/elements/1.1/" version="2.0">
<channel>
<title>This is a test</title>
<link>http://somelink.html</link>
<description>RSS Feed</description>
<item>
<title>This is a title</title>
<link>http://somelink.html</link>
<description><div style='font-size: 9px;'><div class="rendering rendering_researchoutput rendering_researchoutput_short rendering_contributiontojournal rendering_short rendering_contributiontojournal_short"><h2 class="title"><a class="link" rel="ContributionToJournal" href="http://somelink.html"><span>This is a Title</span></a></h2><a class="link person" rel="Person" href="somelink.html"><span>Bob, C. R</span></a> &amp; Smith, W. <span class="date">2014</span> <span class="journal">In : <a class="link" rel="Journal" href="http://somelink.html"><span>Publishers title</span></a>.</span><p class="type"><span class="type_family">Research output<span class="type_family_sep">: </span></span><span class="type_classification_parent">Contribution to journal<span class="type_parent_sep"> › </span></span><span class="type_classification">Article</span></p></div><div class="rendering rendering_researchoutput rendering_researchoutput_detailsportal rendering_contributiontojournal rendering_detailsportal rendering_contributiontojournal_detailsportal"><div class="article"><table class="properties"><tbody><tr class="language"><th>Original language</th><td>English</td></tr><tr><th>Journal</th><td><a class="link" rel="Journal" href="http://somelink.html"><span>Journal of Human Rights and the Environment </span></a></td></tr><tr><th>Journal publication date</th><td>2014</td></tr><tr class="status"><th>State</th><td>In press</td></tr></tbody></table></div></div></div></description>
<pubDate>Wed, 02 Apr 2014 15:59:41 GMT</pubDate>
<guid>http://somelink.html</guid>
<dc:date>2014-04-02T15:59:41Z</dc:date>
</item>
</channel>
</rss>
这是一个测试
http://somelink.html
RSS源
这是一个标题
http://somelink.html
div style='font-size:9px;'div class=“渲染渲染”\u researchoutput渲染”\u researchoutput\u short渲染”\u contributionJournal渲染”\u short渲染”\u contributionJournal\u short“h2 class=“title”a class=“link”rel=“contributionJournal”href=”http://somelink.html“span这是一个标题/span/a/h2a class=“link person”rel=“person”href=“somelink.html”spanBob,C.R/span/a&;amp;史密斯,W.span class=“date”2014/span class=“journal”In:a class=“link”rel=“journal”href=”http://somelink.html“span出版商标题/span/a./spanp class=“type”span class=“type\u family”研究成果span class=“type\u family\u sep”:/span/span/span class=“type\u classification\u parent”对期刊的贡献span class=“type\u parent\u sep”›/span/span class=“type_classification”Article/span/p/divdiv class=“rendering rendering\u researchoutput rendering\u researchoutput\u detailsportal rendering\u contributionJournal rendering\u detailsportal\u contributionJournal\u detailsportal”div class=“Article”table class=“properties”tbodytr class=“language”原始语言/thtdEnglish/td/trthjournal/thtda class=“link”rel=“Journal”href=”http://somelink.html“span人权与环境杂志/span/a/td/TRTHJournal出版日期/thtd2014/td/trtrtr class=“status”thState/thtdIn press/td/tr/tbody/table/div/div
2014年4月2日星期三15:59:41 GMT
http://somelink.html
2014-04-02T15:59:41Z
并演示如何使用XSLT解析
标记以返回
字段或
字段的内容
我在xslt中尝试了以下操作:
<xsl:value-of select="span[@class='date']"/>
它不返回任何内容这里有一种笨拙的方法来提取
元素的内容(或者更确切地说,禁用转义后的
元素是什么):
“如何使用XSLT解析
标记”您不能使用XSLT解析
标记,因为它不是XML。另见:
<xsl:value-of select="substring-before(substring-after(description, '<span class="date">'), '</span>')"/>