Re: [xsl] Testing implicit XHTML hierarchy

["bryan rasmussen" <rasmussen.bryan@xxxxxxxxx>]

well i agree there is redundancy (damn I should really do this stuff
instead of messing around )

but

<xsl:variable name="h"
select="//xhtml:*[matches(local-name(),'^h[1-6]')]/number(substring(local-name(),2))"/>

<xsl:for-each select="1 to count($h)-1">
<result select="$h[.+1][1]">

<xsl:value-of select="$h[.+1][1]"/>
</result>
<result select="$h[.][1]">

<xsl:value-of select="$h[.][1]"/>
</result>
<result select="$h[.+1][2]">
<xsl:value-of select="$h[.+1][2]"/>
</result>
<result select="$h[.][2]"><xsl:value-of select="$h[.][2]"/>
</result>
</xsl:for-each>
</output>
</xsl:template>

produces

<?xml version="1.0" encoding="UTF-8"?>
<output>
  <result select="$h[.+1][1]">2</result>
  <result select="$h[.][1]">1</result>
  <result select="$h[.+1][2]">5</result>
  <result select="$h[.][2]">2</result>
  <result select="$h[.+1][1]">2</result>
  <result select="$h[.][1]">1</result>
  <result select="$h[.+1][2]">5</result>
  <result select="$h[.][2]">2</result>
  <result select="$h[.+1][1]">2</result>
  <result select="$h[.][1]">1</result>
  <result select="$h[.+1][2]">5</result>
  <result select="$h[.][2]">2</result>
  <result select="$h[.+1][1]">2</result>
  <result select="$h[.][1]">1</result>
  <result select="$h[.+1][2]">5</result>
  <result select="$h[.][2]">2</result>
  <result select="$h[.+1][1]">2</result>
  <result select="$h[.][1]">1</result>
  <result select="$h[.+1][2]">5</result>
  <result select="$h[.][2]">2</result>
  <result select="$h[.+1][1]">2</result>
  <result select="$h[.][1]">1</result>
  <result select="$h[.+1][2]">5</result>
  <result select="$h[.][2]">2</result>
  <result select="$h[.+1][1]">2</result>
  <result select="$h[.][1]">1</result>
  <result select="$h[.+1][2]">5</result>
  <result select="$h[.][2]">2</result>
  <result select="$h[.+1][1]">2</result>
  <result select="$h[.][1]">1</result>
  <result select="$h[.+1][2]">5</result>
  <result select="$h[.][2]">2</result>
</output>

which I'm not following how $h[.+1][2] will always be empty?

Since there are two items in the sequence I mean. Sorry if I'm not
very clear right now, but I am not thinking especially sharpish.

Cheers,
Bryan Rasmussen

On 7/5/07, David Carlisle <davidc@xxxxxxxxx> wrote:
> > actually  theres more than one item in your sequence that way.
> I may have mistyped, but you shouldn't need the [1]'s  in for example
> $h[.+1][1]
> as $h[.+1] is %h[position()=.+1] so can return at most 1 item, so [1] is
> redundant.
> similarly $h[.+1][2] will always be empty.
>
> David
>
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