Convert an xml element whose content is inside CDA

2019-01-09 13:39发布

问题:

I have a xml fragment like below

<Detail uid="6">
    <![CDATA[
    <div class="heading">welcome to my page</div>
    <div class="paragraph">this is paraph</div>
    ]]>
</Detail>

and I want to be able to change the

<div class="heading">...</div> to <h1>Welcome to my page</h1>
<div class="paragraph">...</div> to <p>this is paragraph</p>

do you know how I can do that in xslt 1.0

回答1:

What about running two transforms.

Pass 1.)

<?xml version="1.0" encoding="UTF-8"?>
  <xsl:stylesheet
   version="1.0"
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes" encoding="UTF-8"/>

  <xsl:template match="/">
    <xsl:apply-templates />
  </xsl:template>

    <xsl:template match="Detail">
        <Detail>
            <xsl:copy-of select="@*"/>
        <xsl:value-of select="." disable-output-escaping="yes" />
        </Detail>
    </xsl:template>

</xsl:stylesheet>

Will produce:

<?xml version="1.0" encoding="UTF-8"?>
<Detail uid="6"> 
    <div class="heading">welcome to my page</div>
    <div class="paragraph">this is paraph</div>
</Detail>

Pass 2.)

<?xml version="1.0" encoding="UTF-8"?>
  <xsl:stylesheet
   version="1.0"
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes" encoding="UTF-8"/>

  <xsl:template match="/">
    <xsl:apply-templates />
  </xsl:template>

    <xsl:template match="@*|node()">
        <xsl:copy>
            <xsl:apply-templates select="@*| node()" />
        </xsl:copy>
    </xsl:template>

    <xsl:template match="div[@class='heading']">
        <h1><xsl:value-of select="."/></h1>
    </xsl:template>

    <xsl:template match="div[@class='paragraph']">
        <p><xsl:value-of select="."/></p>
    </xsl:template>

</xsl:stylesheet>

Produces:

<?xml version="1.0" encoding="UTF-8"?>
<Detail uid="6">
<h1>welcome to my page</h1>
<p>this is paraph</p>
</Detail>


回答2:

You cannot tell XSL 1.0 to fish a string out of a CDATA and parse it as XML.



回答3:

You can't "remove" the CDATA, but you can achieve the desired output somewhat crudely:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
   <Detail>
        <xsl:variable name="before" select="substring-before(//Detail,'&lt;div class=&quot;heading&quot;&gt;')" />
        <xsl:variable name="afteropen" select="substring-after(//Detail,'&lt;div class=&quot;heading&quot;&gt;')" />
        <xsl:variable name="body" select="substring-before($afteropen, '&lt;/div&gt;')" />
        <xsl:variable name="after" select="substring-after($afteropen, '&lt;/div&gt;')" />
        <xsl:value-of select="concat($before, '&lt;h1&gt;', $body, '&lt;/h1&gt;',$after)"
                disable-output-escaping="yes"       />
   </Detail>
</xsl:template>
</xsl:stylesheet>

This will work for the first type of div you're trying to parse and you can follow something similar with the second one. It could be made more generic with some effort.



标签: xml xslt cdata