转换包含CDATA内容的xml元素

I have a xml fragment like below

我有一个xml片段，如下所示

<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

我想要改变

<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

你知道我如何在xslt 1.0中做到这一点吗

3 个解决方案

#1

What about running two transforms.

运行两个变换怎么样?

Pass 1.)

通过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:

会产生:

<?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.)

通过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:

生产:

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

#2

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

不能告诉XSL 1.0从CDATA中提取字符串并将其解析为XML。

#3

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

您不能“删除”CDATA，但是您可以粗略地实现所需的输出:

<?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.

这将适用于您试图解析的第一种类型的div，您可以遵循与第二种类似的内容。通过一些努力，它可以变得更加通用。

#1