yet another class take-home quiz I need help with to compare the position() return value against a parameter value as follows using XSLT 1.0. I have this simple XML below...
<Providers>
<Company>
<Name>Alpha</Name>
</Company>
<Company>
<Name>Beta</Name>
</Company>
<Company>
<Name>Omega</Name>
</Company>
</Providers>
And I want to pass in a parameter with a numeric value designating which node level to insert a new element called Rating like this...
<Providers>
<Company>
<Name>Alpha</Name>
</Company>
<Company>
<Name>Beta</Name>
<Rating>Good</Rating>
</Company>
<Company>
<Name>Omega</Name>
</Company>
</Providers>
Here's my XSLT...
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:param name="nodeNumber">2</xsl:param>
<xsl:template match="*">
<xsl:copy>
<xsl:copy-of select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="Providers">
<xsl:copy>
<xsl:apply-templates />
</xsl:copy>
</xsl:template>
<xsl:template match="Company[ position() = 2 ]">
<xsl:copy>
<xsl:copy-of select="@*|node()"/>
<Rating>Good</Rating>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Instead of hard-coding to the second node...
<xsl:template match="Company[ position() = 2 ]">
I want to do a comparison like this so it uses a parameter value...
<xsl:template match="Company[ position() = $nodeNumber ]">
but I get an error if XSLT is 1.0...
c:\ msxsl.exe providers.xml providers.xsl
Error occurred while compiling stylesheet 'providers.xsl'.
Code: 0x80004005
Variables may not be used within this expression.
Company[ position() = -->$nodeNumber <--]
it works fine if XSLT is 2.0...
c:\ altovaxml.exe /xslt2 providers.xsl /in providers.xml
<?xml version="1.0" encoding="UTF-8"?>
<Providers>
<Company>
<Name>Verizon</Name>
</Company>
<Company>
<Name>Sprint</Name>
<Rating>Good</Rating>
</Company>
<Company>
<Name>ATT</Name>
</Company>
</Providers>
I need to use XSLT 1.0. Thank you very much for your time and effort...