XSLT output html tags from XML

Question

I have an XML document similar to:

<tag>
   <content>adsfasdf<b>asdf</b></content>
</tag>

I would like for the XSLT to select the content element and show all of the content:

<xsl:value-of select="/tag/content"/> 

The XSLT is configured to render as HTML. Is there a way that I can get the value-of/copy-of to display the exact content without having to render it?

What I'm looking for is

asdfasdf<b>asdf</b> 

And not:

asdfasdf asdf

Answer 1

You need to escape the tag names within the content, I'd recommend something like:

<xsl:template match="content//*">
    <xsl:value-of select="concat('&lt;',name(),'&gt;')"/>
    <xsl:apply-templates/>
    <xsl:value-of select="concat('&lt;/',name(),'&gt;')"/>
</xsl:template>

which you then can call with:

<xsl:apply-templates select="/tag/content"/>

Answer 2

Quick and dirty way:

<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="html" indent="yes"/>

    <xsl:template match="content">
        <xsl:copy-of select="text() | *"/>
    </xsl:template>
</xsl:stylesheet>

Result against your sample will be:

adsfasdf<b>asdf</b>

Another approach:

<xsl:stylesheet version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="html" indent="yes"/>

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