RE: [xsl] FW: Xslt that worked for fop0.20.2 no longer works at later versions

["Hesselberth, Jan" <Jan.Hesselberth@xxxxxxxxxxxxxxxxxx>]

I thought this solved the problem but I have found when using xsl:sort,
the positions reference the unsorted nodes.

I have
		<xsl:variable name="nodes"
select="//property[@type='MoveAssetSuccess']"/>
		<xsl:for-each select="$nodes">
		<xsl:sort select="./property[@name='sourceSPV']"
data-type="text" order="ascending"/>
		<xsl:sort select="./property[@name='targetSPV']"
data-type="text" order="ascending"/>
		<xsl:sort
select="concat(substring(property[@name='mfAccNo'],1,5),substring(proper
ty[@name='mfAccNo'],8,5),substring(property[@name='mfAccNo'],6,2))"/>
		<xsl:variable name="p" select="position()"/>

How can I get the variable nodes to contain the sorted nodes?

Regards
Jan

-----Original Message-----
From: David Carlisle [mailto:davidc@xxxxxxxxx]
Sent: 29 April 2008 09:28
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: [xsl] FW: Xslt that worked for fop0.20.2 no longer works at
later versions



> Unfortunately I don't have time to do that now. Can you tell
> me how I can compare the previous value in a for-each loop

In general that is not possible. Remember  that there is no requirement
that the items are evaluated in any order, and in particular they may be
evaluated in parallel, that's the whole point of declarative side
effect free programming style as used by XSLT.

If you know that the for-each is iterating through all the child  of
some common parent then the previous sibling will is
preceding-sibling::*[1]
but in general if the for-each is iterating over an arbitrary expression
then if you need to access the whole sequence then you may have to
access it something like

<xsl:variable name="nodes" select="some xpath"/>
<xsl:for-each select="$nodes">
<xsl:variable name="p" select="position()"/>

this node <xsl:apply-templates select="."/>

last node <xsl:apply-templates select="$nodes[$p - 1]"/>



David

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