Antwort: Re: Grouping hierarchy path elements

Hi Jeni,

wow, this is great! Thank you so much for that. It almost works perfectly.

Transforming the real document - not the example I posted - I get a stack
overflow error.

So far, what I have found about the error is this:
It always occurs at the same spot, ie. it is reproducible. At the spot it
happens, the path name of the following item is equal to the current item's
path, but appended some more characters. From that I conclude that the
error has its cause at some place the starts-with function was used.
I will dig into that now. Again, thanks a lot for your help.

Mit freundlichen Gr|_en / Sincerely

Daniel Geske


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                      Jeni Tennison
                      <jeni@jenitenniso        An:       "Daniel Geske"
<Daniel.Geske@xxxxxx>
                      n.com>                   Kopie:
xsl-list@xxxxxxxxxxxxxxxxxxxxxx
                                               Thema:    Re:  Grouping
hierarchy path elements
                      2004-08-23 15:04
                      Bitte antworten
                      an Jeni Tennison






Hi Daniel,

> I've been working on an XSLT for days now, and cannot find the solution
to
> one problem. I have a source XML document containing one level of
elements.
> Each element contains a path element that holds information on what path
> the element was extracted from.
>
> Now, through XSL transformation, I would like to recreate the tree
> structure of the items.

Here's a set of templates that works with your example.

The main template is the transformDocument template. This takes a path
(an initial part of a path) and a set of items (whose paths should all
start with the $path). It works out the next step in the path for the
first item and from that creates a new path. Then it sorts the items
into three groups:

  - items whose path *is* the new path, which should just be output
  - items whose path *starts with* the new path, which need to be
    processed again by this template, with the new path
  - items whose path *doesn't* start with the new path, which need to
    be processed again by this template, with the current path

The result of the first two of these groups gets put within a <path>
element, and the result of the third of these groups gets inserted
afterwards.

<xsl:template name="transformDocument">
  <xsl:param name="path"/>
  <xsl:param name="items" select="/.."/>
  <xsl:if test="$items">
    <xsl:variable name="step">
      <xsl:variable name="rest" select="substring-after($items[1]/path,
concat($path, '\'))" />
      <xsl:choose>
        <xsl:when test="contains($rest, '\')">
          <xsl:value-of select="substring-before($rest, '\')" />
        </xsl:when>
        <xsl:otherwise>
          <xsl:value-of select="$rest" />
        </xsl:otherwise>
      </xsl:choose>
    </xsl:variable>
    <xsl:variable name="newPath" select="concat($path, '\', $step)" />
    <path name="{$step}">
      <xsl:apply-templates select="$items[path = $newPath]"/>
      <xsl:call-template name="transformDocument">
        <xsl:with-param name="path" select="$newPath" />
        <xsl:with-param name="items" select="$items[starts-with(path,
$newPath) and
                                                    path != $newPath]"/>
      </xsl:call-template>
    </path>
    <xsl:call-template name="transformDocument">
      <xsl:with-param name="path" select="$path"/>
      <xsl:with-param name="items" select="$items[not(starts-with(path,
$newPath))]"/>
    </xsl:call-template>
  </xsl:if>
</xsl:template>

The next template matches the <items> element and starts off the
processing. I've assumed that there's only one root here.

<xsl:template match="items">
  <transformedDocument>
    <xsl:variable name="root" select="substring-before(item[1]/path, '\')"
/>
    <path name="{$root}">
      <xsl:call-template name="transformDocument">
        <xsl:with-param name="path" select="$root" />
        <xsl:with-param name="items" select="item" />
      </xsl:call-template>
    </path>
  </transformedDocument>
</xsl:template>

The final template just outputs whatever you want for each item; here,
a copy of the <item> element without its child <path> element.

<xsl:template match="item">
  <item id="{@id}" />
</xsl:template>

Cheers,

Jeni

---
Jeni Tennison
http://www.jenitennison.com/