navigating a sorted list

[Khayrat]

Hi folks, please help

I have a xml file with a list of items. The list is sorted during xslt
processing. Based on this sorted list I want a navigation facility to
walk through the sorted list starting from a choosen item via <back> and
<next>.

Say this is the list:

<a>
<b id="1">2</b>
<b id="15">1</b>
<b id="19">45</b>
<b id="1299">10</b>
<b id="121">5</b>
</a>
This is the position of the choosed item, which should be in this case
the third item of the sorted list (e.g. <b id="121">5</b>):

<xsl:variable name="pos">3</xsl:variable>
Here is a scetch of the intended navagation generator:

<xsl:template match="a">
<xsl:for-each select="b">
<xsl:sort/>
<xsl:if test="postion() = $pos">
<xsl:message>
back id: <xsl:value-of select="../b[$pos-1]/@id"/>, pos:
<xsl:value-of select="$pos-1"/>
current id: <xsl:value-of select="self::n ode()/@id"/>, pos:
<xsl:value-of select="$pos"/>
next id: <xsl:value-of select="../b/[$pos+1]/@id"/>, pos:
<xsl:value-of select="$pos+1"/>
</xsl:message>
</xsl:if>
...
</xsl:for-each>
...
</xsl:template>

I also tried with sibling selection:
back id: <xsl:value-of select="precedi ng-sibling::node()/@id"/>, pos:
<xsl:value-of select="$pos-1"/>

But I alway got the ids of original xml list and not the sorted one.

Any help appreciated.

Regards
Khayrat

[Martin Honnen]

Khayrat wrote:

But I alway got the ids of original xml list and not the sorted one.

Consider doing a two step transformation where the first stylesheet
sorts the original XML and the second selects based on that parameter,
taking the result of the first transformation as the input.
Or sort that input into a variable and use an extension function to
convert the variable value (rtf, result tree fragment) into a node set
to process if further.

--

Martin Honnen
http://JavaScript.FAQTs.com/