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RE: [xsl] Easy question, big headache.

2008-03-11 01:33:55

And I need to output some of these records, but not all of 
them. The rules upon which I decide to output are complex, 
and I won't bore you with the details.

Unfortunately it all depends on the detail.

If you can express the rules in an XPath expression EXP, then you can do:

<xsl:for-each select="record[EXP]">
  <xsl:copy-of select="."/>
  <xsl:if test="position() mod 5 = 0">
    <marker/>
  </xsl:if>
</xsl:for-each>

In XSLT 2.0 this always works - because EXP can call a stylesheet function
that can do any computation you like. In 1.0 it rather depends on EXP.

In any other language, it's easy easy:


Writing in a language you know is always easier than in one that's new!

Michael Kay
http://www.saxonica.com/


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