Parsing text as is for selected nodes in XSLT - xslt

Question

Input XML is this:
<input>
<foo>John&apos;s bar</foo>
<bar>test</bar>
<foobar>testing</foobar>
</input>
After XSL transformation:
<input>
<foo>John's bar</foo>
<bar>this_test</bar>
</input>
But the legacy system expects:
<foo>John&apos;s bar</foo>
not <foo>John's bar</foo>
So I want to retain the value under <foo> as is rather than let XSLT parse it.
I tried using <xsl:output method="text"/> but with no luck of success..
I think XML itself when loaded gets parsed and XSLT just outputs as is..
If that's true I atleast want to escape it and make &apos; irrespective of whether it was &apose or ' in the input XML.
XSLT that I tried is this:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format">
<xsl:output method="xml"/>
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="bar">
<xsl:copy>
<xsl:text>this_</xsl:text>
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>
<xsl:template match="foobar"/>
</xsl:stylesheet>

Answer 1

If you are limited to XSLT 1.0, use disable-output-escaping="yes". This attribute can be used on xsl:text and xsl:value-of elements and it is deprecated in XSLT 2.0.
Stylesheet (XSLT 1.0)
<?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"/>
<xsl:strip-space elements="*"/>
<xsl:variable name="vApos">'</xsl:variable>
<xsl:variable name="vAmp">&</xsl:variable>
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="bar">
<xsl:copy>
<xsl:text>this_</xsl:text>
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>
<xsl:template match="foobar"/>
<xsl:template match="foo">
<xsl:variable name="rep">
<xsl:call-template name="replace-string">
<xsl:with-param name="text" select="."/>
<xsl:with-param name="replace" select="$vApos" />
<xsl:with-param name="with" select="concat($vAmp,'apos;')"/>
</xsl:call-template>
</xsl:variable>
<xsl:copy>
<xsl:value-of select="$rep" disable-output-escaping="yes"/>
</xsl:copy>
</xsl:template>
<xsl:template name="replace-string">
<xsl:param name="text"/>
<xsl:param name="replace"/>
<xsl:param name="with"/>
<xsl:choose>
<xsl:when test="contains($text,$replace)">
<xsl:value-of select="substring-before($text,$replace)"/>
<xsl:value-of select="$with"/>
<xsl:call-template name="replace-string">
<xsl:with-param name="text" select="substring-after($text,$replace)"/>
<xsl:with-param name="replace" select="$replace"/>
<xsl:with-param name="with" select="$with"/>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$text"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>
The XSLT 1.0 solution makes use of advice given by Dimitre Novatchev here and Mads Hansen's answer here.
The XSLT 2.0 solution is more elegant, use a character-map to control the serialization of output. Make sure you escape the ampersand character as well (&apos; instead of &apos;).
Stylesheet (XSLT 2.0)
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes" use-character-maps="apo"/>
<xsl:strip-space elements="*"/>
<xsl:character-map name="apo">
<xsl:output-character character="&apos;" string="&apos;"/>
</xsl:character-map>
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="bar">
<xsl:copy>
<xsl:text>this_</xsl:text>
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>
<xsl:template match="foobar"/>
</xsl:stylesheet>
Output (Saxon 9.5 for 2.0, Xalan 2.7.1 for 1.0)
<?xml version="1.0" encoding="UTF-8"?>
<input>
<foo>John&apos;s bar</foo>
<bar>this_test</bar>
</input>