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RE: [xsl] About and

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FAQ,

> In the next xsl I want to make that the param "cod" be 
> incremented in each time that the <xsl:for-each> cicle 
> occurs, but it's result is an error:
> 
> "javax.xml.transform.TransformerException: xsl:with-param is 
> not allowed in this position in the stylesheet!"
> 
> Why?!
> Tks
> 
> <?xml version="1.0" encoding="ISO-8859-1"?>
> <xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
>     version="1.0"
>     xmlns:xalan="http://xml.apache.org/xslt">
> 	<xsl:param name="cod">0</xsl:param>
> 	<xsl:output method="html"/>
> 	<xsl:template match="/">
>     		<xsl:for-each select=".//Boy">
>     			<xsl:with-param name="cod" 
> select="number($cod+1)"/>
> 			<!--make other things-->
> 		</xsl:for-each>
> 	</xsl:template>
> </xsl:transform>

XSLT doesn't allow you to update parameter values after they've been bound. Repeat "recursion is my friend" ten times, and you'll get the hang of it.

Anyhow, rewrite your stylesheet to

  <xsl:param name="cod" select="0" />
  <xsl:template match="/">
    <xsl:for-each select="//Boy">
      <xsl:variable name="_cod" select="$cod + position()" />
	<!--make other things-->
    </xsl:for-each>
  </xsl:template>

And use $_cod instead of $cod in "make other things". Use select attribute in the xsl:param to bind the parameter into a number, instead of a RTF <http://www.w3.org/TR/xslt#section-Result-Tree-Fragments>.

Cheers,

Jarno

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