Is it possible to replace & with & in XSLT? - xslt

Question

I tried to do that with replace($val, 'amp;', ''), but seems like & is atomic entity to the parser. Any other ideas?
I need it to get rid of double escaping, so I have constructions like ᾰ in input file.
UPD:
Also one important notice: I have to make this substitution only inside of specific tags, not inside of every tag.

Answer 1

If you serialize there is always (if supported) the disable-output-escaping hack, see http://xsltransform.hikmatu.com/nbUY4kh which transforms
<root>
<foo>a & b</foo>
<bar>a & b</bar>
</root>
selectively into
<root>
<foo>a & b</foo>
<bar>a & b</bar>
</root>
by using <xsl:value-of select="." disable-output-escaping="yes"/> in the template matching foo/text():
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="foo/text()">
<xsl:value-of select="." disable-output-escaping="yes"/>
</xsl:template>
</xsl:transform>
To achieve the same selective replacement with a character map you could replace the ampersand in foo text() children (or descendants if necessary) with a character not used elsewhere in your document and then use the map to map it to an unescaped ampersand:
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:output use-character-maps="doe"/>
<xsl:character-map name="doe">
<xsl:output-character character="«" string="&"/>
</xsl:character-map>
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="foo/text()">
<xsl:value-of select="replace(., '&', '«')"/>
</xsl:template>
</xsl:transform>
That way
<root>
<foo>a & b</foo>
<bar>a & b</bar>
</root>
is also transformed to
<root>
<foo>a & b</foo>
<bar>a & b</bar>
</root>
see http://xsltransform.hikmatu.com/pPgCcoj for a sample.

Answer 2

If your XML contains ᾰ and you believe that this is a double-escaped representation of the character with codepoint 8112, then you can convert it to this character using the XPath expression
codepoints-to-string(xs:integer(replace($input, '&#([0-9]+);', $1)))
remembering that if you write this XPath expression in XSLT, then the & must be written as &.