On Thu, Jun 11, 2009 at 1:45 AM, David
Carlisle<davidc(_at_)nag(_dot_)co(_dot_)uk> wrote:
Another xslt1 version (as for Dimitre's this works also in xslt2 as
written, with version="1.0" but if you make it version="2.0" and turn off
backward compatibilty mode, you get a type error).
Yes, both David's solution and mine cause a type error to be raised if
they use version="2.0" and an XSLT 2.0 processor.
This one performs with both versions without any error:
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="text"/>
<xsl:variable name="vrtfResults">
<r>1</r>
<r>1.70</r>
<r>2.25</r>
<r>3</r>
</xsl:variable>
<xsl:param name="optional.one" select="true()" />
<xsl:param name="optional.two" select="true()" />
<xsl:param name="optional.three" select="true()" />
<xsl:template match="/">
<xsl:variable name="vOne" select=
"document('')/*/xsl:variable[(_at_)name='vrtfOne']"
/>
<xsl:variable name="vResults" select=
"document('')/*/xsl:variable[(_at_)name='vrtfResults']/*"
/>
<xsl:value-of select=
"$vResults[number($optional.one)
+ number($optional.two)
+ number($optional.three)
+ 1
]"/>
</xsl:template>
</xsl:stylesheet>
--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.
---------------------------------------
To invent, you need a good imagination and a pile of junk
-------------------------------------
Never fight an inanimate object
-------------------------------------
You've achieved success in your field when you don't know whether what
you're doing is work or play
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