How to call named templates based on a variable?

Question

I don't know if it's possible, but I'm wondering how to do it...

Let's say we have the following XSL:

<xsl:template name="foo">
  Bla bla bla
</xsl:template>
...
<xsl:template name="bar">
  Bla bla bla
</xsl:template>
...
<xsl:template match="/">
  <xsl:if test="$templateName='foo'">
    <xsl:call-template name="foo"/>
  </xsl:if>
  <xsl:if test="$templateName='bar'">
    <xsl:call-template name="bar"/>
  </xsl:if>
</xsl:template>

Is it possible to change the XSL to read something like...

<xsl:template match="/">
  <xsl:call-template name="$templateName"/>
</xsl:template>

Answer 1

It's not possible exactly as you describe, but if you want to be able to choose a template at run-time based on some value you set elsewhere, there is a trick to do that. The idea is to have your named template also match a node with a corresponding name in a distinct mode (so that it doesn't mess up your normal transformation), and then match on that. For example:

<xsl:stylesheet ... xmlns:t="urn:templates">

  <!-- Any compliant XSLT processor must allow and ignore any elements 
       not from XSLT namespace that are immediate children of root element -->
  <t:templates>
    <t:foo/>
    <t:bar/>
  </t:templates>

  <!-- document('') is the executing XSLT stylesheet -->     
  <xsl:variable name="templates" select="document('')//t:templates" />

  <xsl:template name="foo" match="t:foo" mode="call-template">
    Bla bla bla
  </xsl:template>

  <xsl:template name="bar" match="t:foo" mode="call-template">
    Bla bla bla
  </xsl:template>

  <xsl:template match="/">
    <xsl:variable name="template-name" select="..." />
    <xsl:apply-templates select="$templates/t:*[local-name() = $template-name]"
                         mode="call-template"/>
  </xsl:template>

Note that you can use <xsl:with-param> in <xsl:apply-templates>, so you can do everything with this that you could do with a plain <xsl:call-template>.

Also, the code above is somewhat lengthier than you might need because it tries to avoid using any XSLT extensions. If your processor supports exslt:node-set(), then you can just generate nodes directly using <xsl:element>, and use node-set() to convert the resulting tree fragment to a plain node to match against, without the need for document('') hack.

For more information, see FXSL - it's a functional programming library for XSLT that is based on this concept.

Answer 2

No, this is not possible not directly possible. The calling convention is:

<xsl:call-template name="QName" />

Where a QName is defined as:

QName ::= PrefixedName | UnprefixedName

PrefixedName   ::= Prefix ':' LocalPart
UnprefixedName ::= LocalPart

Prefix         ::= NCName
LocalPart      ::= NCName

Basically this boils down to "characters only, no expressions". As the other answers highlight, there are in fact ways to do something equivalent, but the straightforward approach/naïve approach will not work.

Answer 3

For anyone's future reference:

Here is a working example based on Pavel Minaev's answer. No original thought on my part. ;-) I switched it to use msxml:node-set as he described (more or less) so that it works in .NET.

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" version="1.0">
    <xsl:variable name="templates">
        <templates>
            <foo />
            <bar />
        </templates>
    </xsl:variable>
    <xsl:template name="foo" match="foo" mode="call-template">
        <FooElement />
    </xsl:template>
    <xsl:template name="bar" match="bar" mode="call-template">
        <BarElement />
    </xsl:template>
    <xsl:template match="/">
        <Root>
            <xsl:variable name="template-name">bar</xsl:variable> <!-- Change this to foo to get the other template. -->
            <xsl:apply-templates select="msxsl:node-set($templates)/*/*[local-name() = $template-name]" mode="call-template" />
        </Root>
    </xsl:template>
</xsl:stylesheet>