Re: [xsl] getting the node position in source xml in a variable

[Jeni Tennison <jeni@xxxxxxxxxxxxxxxx>]

Hi Gurvinder,

I'm afraid that I can't tell what's going wrong without you supplying
some more information, specifically about how you're calling the
template. I tried using this source XML:

<test>
  <datanodes>
    <data>3</data>
    <data>2</data>
    <data>1</data>
  </datanodes>
  <displaynodes>
    <display>C</display>
    <display>B</display>
    <display>A</display>
  </displaynodes>
</test>

With this call to the template:

  <xsl:call-template name="lookup">
    <xsl:with-param name="name" select="'test'" />
    <xsl:with-param name="datanodes"
                    select="/test/datanodes/data" />
    <xsl:with-param name="displaynodes"
                    select="/test/displaynodes/display" />
  </xsl:call-template>

And I got this result:

<select name="test">
  <option id="1" value="1">1-A</option>
  <option id="2" value="2">2-B</option>
  <option id="3" value="3">3-C</option>
</select>

One problem that could be occurring is if the $displaynodes aren't
actually siblings of each other. But without seeing how you're
actually calling the template and what the source of your stylesheet
looks like, I can't tell.

Cheers,

Jeni

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


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list