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[@s=current()/@ix]"/>
>    <xsl:text>&#xa;</xsl:text>
>  </xsl:template>
>  <xsl:template match="ln">
>    <xsl:value-of select="concat(' ', @d)"/>
>    <xsl:apply-templates select="following-sibling::ln[@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.