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At 2003-02-24 15:14 +0200, Jarno.Elovirta@xxxxxxxxx wrote:
Note, Jarno, the sorting that the original poster is doing.
This will visit the source node tree in document order, and not in sorted order.
Unless I'm missing something (easily possible early on a Monday morning), I'm not convinced a recursive template addressing nodes in document order is going to quite address the poster's needs to know the "previous" node in sorted order.
I hope this helps.
..................... Ken
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RE: [xsl] Comparing and re-assigning variables.
Subject: RE: [xsl] Comparing and re-assigning variables. From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx> Date: Mon, 24 Feb 2003 08:29:47 -0500 |
At 2003-02-24 15:14 +0200, Jarno.Elovirta@xxxxxxxxx wrote:
> Ok, here's a part of a xsl file i wrote to create a > transactions report: > > <!-- Iterate over all txns. --> > <table border="0"> > <xsl:for-each select="report-root/txns"> > <xsl:sort select="cpmBI"/> > <xsl:sort select="date"/>
Note, Jarno, the sorting that the original poster is doing.
Using xsl:for-each you don't; if you process the txns using a recursive template construct, you can always pass the previous currCpmBI, but not using xsl:for-each - it's *not* a for-loop like in, say, Java. Instead just do
<xsl:if test="cpmMemBI = previous-sibling::txns[1]/cpmMemBI">
This will visit the source node tree in document order, and not in sorted order.
Unless I'm missing something (easily possible early on a Monday morning), I'm not convinced a recursive template addressing nodes in document order is going to quite address the poster's needs to know the "previous" node in sorted order.
I hope this helps.
..................... Ken
-- Upcoming hands-on in-depth XSLT/XPath and/or XSL-FO North America: June 16-20, 2003
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