Adding special characters to xml output - xslt

Question

We have a piece of code which returns XML in a format like:
Source XML:
<Root>
<Book>
<BookId>a</BookId>
<Description>aDescription</Description>
</Book>
<Book>
<BookId>b</BookId>
<Description>bDescription</Description>
</Book>
</Root>
I want to replace the special characters with the literal characters...
<
will be < etc.
I know I can use:
<xsl:character-map name="escapeMapper">
<xsl:output-character character="<" string="<"/>
<xsl:output-character character=">" string=">"/>
</xsl:character-map>
However here is the twist, I want to convert special characters first, then run the resulting XML through other templates. So, I want to run the source XML through a template replacing the special characters, putting the result in to a variable:
<xsl:variable name="vrtfPass1">
Now I can use the multi-pass technique and apply other templates using the variable as the source.
How can I convert the special characters into the literal characters?

Answer 1

It sounds like what you want to do is not convert < and > to < and >, but to parse the XML, yes? Are you able to use Saxon extension functions in what you're doing? If so, I think you could do something like this:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:saxon="http://saxon.sf.net/">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="/">
<xsl:variable name="loaded">
<xsl:apply-templates />
</xsl:variable>
<xsl:apply-templates select="$loaded/*" />
</xsl:template>
<xsl:template match="#* | node()">
<xsl:copy>
<xsl:apply-templates select="#* | node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Book/text()[normalize-space()]">
<xsl:copy-of select="saxon:parse(.)"/>
</xsl:template>
</xsl:stylesheet>
Though this assumes that the text children of Book would always be well-formed XML in its own right.