This is a recursive query, and is therefore beyond the scope of XPath alone:
it needs recursive functions. But it's easy in XSLT (untested!):
<xsl:key name="k" match="ln" use="@s"/>
<xsl:function name="my:children" as="xs:string*">
<xsl:param name="parent" as="xs:string"/>
<xsl:sequence select="key('k', $parent)/@d"/>
</xsl:function>
<xsl:function name="my:descendants" as="xs:string*">
<xsl:param name="parent" as="xs:string"/>
<xsl:variable name="children" select="my:children($parent)"/>
<xsl:sequence select="$children, for $c in $children return
my:descendants($c)"/>
</xsl:function>
A bit harder if you need to add a check for cycles, but still doable.
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Rolf Schumacher [mailto:mailinglist(_at_)august(_dot_)de]
Sent: 08 May 2007 06:01
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] matching elements of a list
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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