thanks, that makes sense. so it's a matter of serialization
----- Ursprüngliche Mail ----
Von: Martin Honnen <Martin(_dot_)Honnen(_at_)gmx(_dot_)de>
An: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Gesendet: Donnerstag, den 8. Oktober 2009, 11:55:26 Uhr
Betreff: Re: [xsl] copy-of converts hex to decimal
Matthias Müller wrote:
can someone explain why xsl:copy-of converts hex values to decimal?
e.g. the node:
<HexDecTest>
<Hex>Ⅰ</Hex>
<Dez>Ⅰ</Dez>
</HexDecTest>
is processed to:
<Result>
<Hex>Ⅰ</Hex>
<Dez>Ⅰ</Dez>
</Result>
using this template:
<xsl:template match="HexDecTest">
<xsl:element name="Result">
<xsl:copy-of select="Hex"/>
<xsl:copy-of select="Dez"/>
</xsl:element>
</xsl:template>
i expected the hex value unchanged,
XSLT works on the XSLT/XPath data model, a tree model. In that model the Hex
and the Dez elements have a text node with character data of Unicode characters
as a child. The input tree is transformed to a result tree, again with text
nodes of Unicode characters.
That result tree can then be serialized and only then you might get characters
escaped, for instance when you have specified an output encoding in which the
characters can not be directly represented. Your serializer has choosen decimal
numbers for escaping.
So the main point is that XSLT does not operate on the markup representation of
the input and xsl:copy-of certainly is not a way to copy the markup in the
input literally to the output. XSLT operates on a tree often created by parsing
an XML document and then creates a result tree and optionally serializes that
result tree, often as XML.
--
Martin Honnen
http://msmvps.com/blogs/martin_honnen/
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