Hi AlexR,

Sadly, that's not how the node-set() function works. You'll need an XSLT extension for that...

/Chriztian

- Refer to this comment to see how the node-set() function can be used: http://our.umbraco.org/forum/developers/xslt/9360-xsl-if-array?p=0#comment34272

/Chriztian

What function can be used to convert string to node set?

I see that xslt very, very, very bad language.

I asked about 5 simple questions and got only one answer. All other questions don't have answers.

Hi Alez

xslt can be frustrating (unless you're as fluent as Chriztian!) but this should work for you

<xsl:variable name="items">
    <items>
     <item>1</item>
        <item>2</item>
        <item>bucklemyshoe</item>
 </items>
</xsl:variable>

<xsl:variable name="itemSet" select="msxml:node-set($items)" />

<xsl:value-of select="count($itemSet//item)" />
-- OUTPUTS "3" ---

<xsl:value-of select="$itemSet//item[position() = 3]" />
-- OUTPUTS "bucklemyshoe" --

Hope this helps,
Dan

Hi AlexR,

It's not that XSLT is a bad language, (quite the opposite), it more that there is a misunderstanding that it's a declarative language, but its all too often treated (and abused) as a functional/procedural language.

XSLT's power is in transforming XML data-sources, (sounds obvious I know), but when trying to create a string-based XML node-set, I can see how this is frustrating and confusing!  Chriztian and Dan's suggestion of creating the XML nodes in a variable, then applying the msxml:node-set function is the best way of achieving this.

I'm curious what you meant about your "5 simple questions" ... what else is troubling you with XSLT? (we're happy to help)

Cheers, Lee.

I known that code from dandrayne will work, but I have code that ca not be transfered to code like dandrayne's code.

For example I writes xslt template that generate range sequence of integers.

<xsl:template name="rangeListLoop">
  <xsl:param name="start" />
  <xsl:param name="count" />
  <xsl:param name="step" />
  <xsl:param name="itemElement" />

  <xsl:if test="$count > 0">
    <xsl:text>&lt;</xsl:text><xsl:value-of select="$itemElement" /><xsl:text>></xsl:text>
      <xsl:value-of select="$start" />
    <xsl:text>&lt;/</xsl:text><xsl:value-of select="$itemElement" /><xsl:text>></xsl:text>

    <xsl:call-template name="rangeListLoop">
      <xsl:with-param name="start" select="$start + $step" />
      <xsl:with-param name="count" select="$count - 1" />
      <xsl:with-param name="step" select="$step" />
      <xsl:with-param name="itemElement" select="$itemElement" />
    </xsl:call-template>
  </xsl:if>

</xsl:template>

And in this code I used itemElement param to generate integer in different elements, like this:

<item>1<item><item>2<item>  or  <month>1<month><month>2<month> and etc.

But I can not use <element name="$itemElement"> because XSLT forbidden use variables in names. But why? What is wrong with generation not static XML elements.

XSLT created to generate XML and it can not create flexible XML, only static. Why?

I tried to use trick to generate elements as text

<xsl:text>&lt;</xsl:text><xsl:value-of select="$itemElement" /><xsl:text>></xsl:text>

But I can not use it also. Because XSLT can not convert string to node set. Why it can not convert string to node set? What wrong with this?

I so on, so no, so on...

XSLT bad language. It  is not flexible and have many unusual restrictions.

Other big restriction that I can not use variables in xpath expressions. I can not generate xpath as string.

How I can flexible transform XML when XSLT is not flexible, it is static.

You should hard-code all in your code - it is stupid.

So I think that is better to call C# functions that will use LINQ to XML and can generate anything without unusual restrictions.

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