Thank you very much, Jeff.
Worked right away on that sample.
Targeting it to my more complex problem (XMI spec)
is not as easy because the links are not just a sibling.
However with your example I should be able to figure out
how that will work.
Thank you, again
Rolf
Jeff Sese wrote:
try this stylesheet, i think this would work (if i guessed your
requirements right).
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0">
<xsl:template match="/">
<xsl:apply-templates select="root/st"/>
</xsl:template>
<xsl:template match="st">
<xsl:value-of select="concat(@ix,':')"/>
<xsl:apply-templates
select="following-sibling::ln[(_at_)s=current()/@ix]"/>
<xsl:text>
</xsl:text>
</xsl:template>
<xsl:template match="ln">
<xsl:value-of select="concat(' ', @d)"/>
<xsl:apply-templates select="following-sibling::ln[(_at_)s=current()/@d]"/>
</xsl:template>
</xsl:stylesheet>
--
Jeff
Rolf Schumacher wrote:
do I have reached the limits of xpath?
if several linked lists are contained in one document how to match all
nodes belonging to a specific start node?
The following example may illustrate my question:
Input:
<root>
<st ix="a"/>
<st ix="b"/>
<el ix="c"/>
<el ix="d"/>
<el ix="e"/>
<el ix="f"/>
<el ix="g"/>
<ln s="a" d="g"/>
<ln s="g" d="f"/>
<ln s="b" d="e"/>
<ln s="e" d="c"/>
<ln s="c" d="d"/>
</root>
disired output:
a: g f
b: e c d
How to accomplish that by XSLT (2.0)?
Even your answer "I know for sure that there is not elegant way"
or "You got to extend the processor by some Java function"
would help.
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