我需要通过从外部html文件中获取信息来使用xslt转换创建html:
我的输入XML文件:
<topic xmlns:ditaarch="http://dita.oasis-open.org/architecture/2005/" xmlns:r="http://www.Corecms.com/Core/ns/metadata" class="- topic/topic " ditaarch:DITAArchVersion="1.2" domains="(topic hi-d) (topic indexing-d) (topic d4p_formatting-d) a(props d4p_renditionTarget) (topic d4p_math-d) (topic d4p_variables-d) (topic d4p_verse-d) (topic learningInteractionBase2-d learning2-d+learning-d) (topic learningBase+learningInteractionBase-d) (topic learningInteractionBase-d) (topic learningInteractionBase2-d) (topic xml-d) a(base CoreIdAtt) (topic sdClassification-d) a(base contentstore) " id="T1" outputclass="interactive" r:CoreId="4567">
<title class="- topic/title "/>
<body class="- topic/body ">
<bodydiv class="- topic/bodydiv ">
<xref class="- topic/xref " format="html" href="1234.html" scope="external"/>
</bodydiv>
</body>
</topic>
我使用element调用了html文件。相应的 1234.html 文件代码为:
<html>
<head></head>
<body>
<h1>New HTML test with XSL transformation</h1>
<p>this is a new test about rsuite-edit opening HTML files</p>
</body>
</html>
XSL我试过如下:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:xd="http://www.oxygenxml.com/ns/doc/xsl"
xmlns:r="http://www.Corecms.com/Core/ns/metadata" xmlns:exsl="http://exslt.org/common"
xmlns:xhtml="http://www.w3.org/1999/xhtml" xmlns="http://www.w3.org/1999/xhtml"
xmlns:ditaarch="http://dita.oasis-open.org/architecture/2005/"
xmlns:df="http://dita2indesign.org/dita/functions"
exclude-result-prefixes="xs xd r exsl xhtml ditaarch df" version="2.0">
<xsl:output method="xml" indent="yes"/>
<xsl:function name="df:class" as="xs:boolean">
<xsl:param name="elem" as="element()"/>
<xsl:param name="classSpec" as="xs:string"/>
<xsl:variable name="normalizedClassSpec" as="xs:string" select="normalize-space($classSpec)"/>
<xsl:variable name="result"
select="matches($elem/@class, concat(' ', $normalizedClassSpec, ' | ', $normalizedClassSpec, '$'))"
as="xs:boolean"/>
<xsl:sequence select="$result"/>
</xsl:function>
<xsl:template match="/">
<xsl:variable name="html">
<xsl:apply-templates/>
</xsl:variable>
<xsl:copy-of select="$html"/>
</xsl:template>
<xsl:template match="*[df:class(., 'topic/topic')]">
<div>
<xsl:attribute name="contenteditable">true</xsl:attribute>
<xsl:apply-templates select="@*"/>
<xsl:apply-templates/>
<xsl:choose>
<xsl:when test="not(ancestor::*[df:class(., 'topic/topic')])">
<xsl:apply-templates select="." mode="generate-comments"/>
</xsl:when>
</xsl:choose>
</div>
</xsl:template>
<xsl:template match="*[df:class(., 'topic/title')][parent::*[df:class(., 'topic/topic')]]">
<xsl:variable name="headingLevel" select="count(ancestor::*[df:class(., 'topic/topic')])"
as="xs:integer"/>
<xsl:element name="h{$headingLevel}">
<xsl:apply-templates select="@*"/>
<xsl:apply-templates/>
</xsl:element>
</xsl:template>
<xsl:template match="*[df:class(., 'topic/body')]">
<div>
<xsl:apply-templates select="@*"/>
<xsl:apply-templates/>
</div>
</xsl:template>
<xsl:template match="*[df:class(., 'topic/bodydiv')]">
<div>
<xsl:apply-templates select="@*"/>
<xsl:apply-templates/>
</div>
</xsl:template>
<xsl:template match="*[df:class(., 'topic/p')]">
<p>
<xsl:apply-templates select="@*"/>
<xsl:apply-templates/>
</p>
</xsl:template>
<xsl:template match="*[df:class(., 'topic/xref')]">
<xsl:choose>
<xsl:when test=". != ''">
<a>
<xsl:apply-templates select="@*"/>
<xsl:apply-templates/>
</a>
</xsl:when>
<xsl:otherwise>
<a>
<xsl:apply-templates select="@*"/>
NO URL PROVIDED
</a>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>
我期待输出如下:
<div contenteditable="true" data-coreid="4567" html-coreid="1234"> <!-- Where 1234 is the core ID of the HTML file and 4567 is the DITA topic core ID. -->
<article>
<h1>New</h1>
<p>this is a new test</p>
</article>
</div>
我需要获取外部html内容(1234.html)并使用XSLT创建新的xhtml文件。我是XSL的新手。您的帮助将会很明显。提前致谢
答案 0 :(得分:2)
通常document()
用于检索外部节点。外部文档必须是可解析的。样式表中有很多部分,在我看来并没有多大意义,例如。您使用<xsl:apply-templates select="@*"/>
来处理属性,但由于您没有属性模板,因此只会引用其文本节点。
答案 1 :(得分:1)
鉴于 1234.html 是可解析的,如Ferestes的答案所述
<article>
<h1>New</h1>
<p>this is a new test</p>
</article>
你可以使用
行<xsl:copy-of select="document(@href)"/>
在您的*[df:class(., 'topic/xref')]
模板中复制 1234.html (href
属性)文件中格式正确的内容
编辑(与编辑问题有关):
整个模板可能如下所示:
<xsl:template match="*[df:class(., 'topic/xref')]">
<div contenteditable="true" data-coreid="{ancestor::topic/@r:CoreId}" html-coreid="{substring-before(@href,'.')}">
<xsl:copy-of select="document(@href)"/>
</div>
</xsl:template>
<强>输出:强>
<div xmlns="http://www.w3.org/1999/xhtml" contenteditable="true" data-coreid="4567" html-coreid="1234">
<html xmlns="">
<head/>
<body>
<h1>New HTML test with XSL transformation</h1>
<p>this is a new test about rsuite-edit opening HTML files</p>
</body>
</html>
</div>