XSLT将特定节点下的xml块转换为该节点的xml转义内容 [英] XSLT convert xml block under a specific node to xml-escaped content of that node
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问题描述
INPUT：
< p为H.
< div>
原始< br />这是原始的< b>酸/羟基单羧酸< span class =hl1>酸< / span> ;.
< / div>
< / p>
期望的输出：
< p为H.
< div>
原始& lt / br /& gt;这是原始& lt; b& gt;酸& lt; / b& gt;羟基单羧酸& span class = HL1&安培; QUOT;&安培;氨基甲酸&安培; LT /跨度&安培; GT ;.
< / div>
< / p>
尝试1：
`< xsl：stylesheet version =2.0xmlns：xsl =http://www.w3.org/1999/XSL/Transform>
< xsl：output omit-xml-declaration =yesindent =noencoding =UTF-8/>
<！ - 身份模板 - >
< xsl：template match =node（）| @ *>
< xsl：copy>
< xsl：apply-templates select =node（）| @ */>
< / xsl：copy>
< / xsl：template>
< xsl：template match =div>
< xsl：copy>
< xsl：value-of select =/disable-output-escaping =no/>
< / xsl：copy>
< / xsl：template>
`
Attempt2：
作为替代，我想到将子元素的内容放入CDATA包装器中。
< xsl：stylesheet version = 2.0xmlns：xsl =http://www.w3.org/1999/XSL/Transform>
< xsl：output omit-xml-declaration =yesindent =noencoding =UTF-8/>
<！ - 身份模板 - >
< xsl：template match =node（）| @ *>
< xsl：copy>
< xsl：apply-templates select =node（）| @ */>
< / xsl：copy>
< / xsl：template>
< xsl：template match =div>
< xsl：copy>
< xsl：text disable-output-escaping =yes>& lt; lt;！[CDATA [< / xsl：text>
< xsl：value-of select =//>
< xsl：text disable-output-escaping =yes>]]& gt< / xsl：text>
< / xsl：copy>
< / xsl：template>
但是没有给我我想要的是。
任何人有更好的主意？我使用XSLT 2.0
解决方案
这是一个使用XSLT 3.0 serialize（） PE和EE：
< xsl：stylesheet version =3.0xmlns：xsl =http：//www.w3 .org / 1999 / XSL / Transform
xmlns：xs =http://www.w3.org/2001/XMLSchema
exclude-result-prefixes =xs>
< xsl：template match =@ * | node（）>
< xsl：copy>
< xsl：apply-templates select =@ * | node（）/>
< / xsl：copy>
< / xsl：template>
< xsl：template match =div>
< xsl：copy>
< xsl：apply-templates mode =serialize/>
< / xsl：copy>
< / xsl：template>
< xsl：template match =node（）mode =serialize>
< xsl：variable name =ser-params>
< output：serialization-parameters xmlns：output =http://www.w3.org/2010/xslt-xquery-serialization>
< output：omit-xml-declaration value =yes/>
< / output：serialization-parameters>
< / xsl：variable>
< xsl：value-of select =serialize（。，$ ser-params / *）/>
< / xsl：template>
< / xsl：stylesheet>
对于旧版本的Saxon 9，您可以使用扩展函数serialize，如 http://xsltransform.net/pPqsHTx ：
<？xml version =1.0encoding =UTF-8？>
< xsl：stylesheet version =2.0xmlns：xsl =http://www.w3.org/1999/XSL/Transform
xmlns：xs =http：// www。 w3.org/2001/XMLSchema
exclude-result-prefixes =xs>
< xsl：output name =inlineomit-xml-declaration =yes/>
< xsl：template match =@ * | node（）>
< xsl：copy>
< xsl：apply-templates select =@ * | node（）/>
< / xsl：copy>
< / xsl：template>
< xsl：template match =div>
< xsl：copy>
< xsl：apply-templates mode =serialize/>
< / xsl：copy>
< / xsl：template>
< xsl：template match =node（）mode =serialize>
< xsl：value-of xmlns：saxon =http://saxon.sf.net/select =saxon：serialize（。，'inline'）/>
< / xsl：template>
< / xsl：stylesheet>
Any ideas of how the following problem can be solved would be highly appreciated.
INPUT:
<p>
<div>
Original<br/>This is the original <b>acid</b>, a hydroxy monocarboxylic <span class="hl1">acid</span>.
</div>
</p>
Desired Output:
<p>
<div>
Original<br/>This is the original <b>acid</b>, a hydroxy monocarboxylic <span class="hl1">acid</span>.
</div>
</p>
Attempt 1:
`<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >
<xsl:output omit-xml-declaration="yes" indent="no" encoding="UTF-8"/>
<!--The identity template -->
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="div">
<xsl:copy>
<xsl:value-of select="/" disable-output-escaping="no"/>
</xsl:copy>
</xsl:template>
`
Attempt2: as an alternative, I thought of placing the child elements' content into a CDATA wrapper.
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >
<xsl:output omit-xml-declaration="yes" indent="no" encoding="UTF-8"/>
<!--The identity template -->
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="div">
<xsl:copy>
<xsl:text disable-output-escaping="yes"><![CDATA[</xsl:text>
<xsl:value-of select="/" />
<xsl:text disable-output-escaping="yes">]]></xsl:text>
</xsl:copy>
</xsl:template>
But that does not give me what I want. Anyone with a better idea? I'm using XSLT 2.0
解决方案
Here is a suggestion using XSLT 3.0 serialize() as supported by Saxon 9.6 HE, PE and EE:
<xsl:stylesheet version="3.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xs">
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="div">
<xsl:copy>
<xsl:apply-templates mode="serialize"/>
</xsl:copy>
</xsl:template>
<xsl:template match="node()" mode="serialize">
<xsl:variable name="ser-params">
<output:serialization-parameters xmlns:output="http://www.w3.org/2010/xslt-xquery-serialization">
<output:omit-xml-declaration value="yes"/>
</output:serialization-parameters>
</xsl:variable>
<xsl:value-of select="serialize(., $ser-params/*)"/>
</xsl:template>
</xsl:stylesheet>
With older version of Saxon 9 you could use the extension function serialize, as shown in http://xsltransform.net/pPqsHTx:
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xs">
<xsl:output name="inline" omit-xml-declaration="yes"/>
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="div">
<xsl:copy>
<xsl:apply-templates mode="serialize"/>
</xsl:copy>
</xsl:template>
<xsl:template match="node()" mode="serialize">
<xsl:value-of xmlns:saxon="http://saxon.sf.net/" select="saxon:serialize(., 'inline')"/>
</xsl:template>
</xsl:stylesheet>
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