RE: Can anyone point me in the right direction?

> -----Original Message-----
> From: Patricia LaRue
>
> "How do I display only that worksheet section of the
> XML file?"
So (guessing a bit more) the source XML has subelements on the worksheet
nodes (--correct?):

Very simplified typical example

<root>
  <worksheet name="ws_name1">
    <row id="...">
      <column></column>
      <!-- more columns -->
    </row>
    <!-- more rows -->
  </worksheet>
  <!-- more worksheets -->
</root>

But then I'm losing it a bit... You need to display the subelements of the
worksheet nodes below the hyperlink layed-out in an HTML table?

If so, to transform the above into a very simple HTML table placed below the
hyperlink, using three templates:

<xsl:template match="worksheet">

  <a>
    <xsl:attribute name="href">
      <xsl:value-of select="@name" />
    </xsl:attribute>
  </a>

  <table>
  <xsl:apply-templates select="row" />
  </table>

</xsl:template>

<xsl:template match="row">

  <tr><xsl:apply-templates select="column" /></tr>

</xsl:template>

<xsl:template match="column">

  <td><xsl:value-of select="." /></td>

</xsl:template>

> So far, everything I've read says that I need XPath,
> XLink, XPointer, etc.  I've been reading for days on
> this and I'm still confused about how these work
> together or even if they work in the current version.
> Can anyone tell me whether I should be looking at
> XPath and XSLT? or XLink and XPointer? or should I
> break up the XML file into seperate worksheet and XSLT
> files?  I'm just looking for someone to point me in
> the right direction and I'm greatfull for any
> suggestions.
Depends on what exactly you want to do --which you still haven't told us, so
it's difficult to tell... ;)
Anyway, if the above is sufficient, then XSLT/XPath is definitely *the* way
to go here.


Cheerz,

Andreas


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