Re: matching elements of a list

I should learn about key(...) and lists, Michael, in order to understand
how this will work.

However, I've got to check for cycles ...

Michael Kay wrote:
> 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@xxxxxxxxx] 
>> Sent: 08 May 2007 06:01
>> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
>> Subject:  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.