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

[Jeni Tennison <jeni@xxxxxxxxxxxxxxxx>]

Hi Gurvinder,

> I have one more query... you said that i am not passing the result
> tree fragment what does it mean... i thought when i select nodes and
> pass it as parameter is is passing result tree fragment

No - when you select nodes and pass them using the select attribute of
xsl:with-param, then you pass those nodes as a *node set*. It's only
if you use the *content* of the xsl:with-param that you create a
result tree fragment. For example, if you did:

  <xsl:call-template name="lookup">
    <xsl:with-param name="name">test</xsl:with-param>
    <xsl:with-param name="nodes">
      <xsl:copy-of select="/test/Level2" />
    </xsl:with-param>
  </xsl:call-template>

With this call, both $name and $nodes are result tree fragments.

It doesn't matter too much with $name, because it's a result tree
fragment whose root node only has one child - a text node with the
value 'test' - although a result tree fragment is a little more
unweildy for the processor than a plain string would be.

On the other hand, the $nodes parameter is set to a result tree
fragment whose root node has a number of Level2 element children.
Because that result tree fragment has structure, you usually have to
convert it back into a node set (e.g. using msxsl:node-set()) to do
anything useful with it.

This means that in general you should use the select attribute to set
the values of variable-binding elements, such as xsl:variable,
xsl:param and xsl:with-param.

Cheers,

Jeni

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


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