Question

Is there way to call an XSL template with optional parameters?

For example:

<xsl:call-template name="test">
  <xsl:with-param name="foo" select="'fooValue'" />
  <xsl:with-param name="bar" select="'barValue'" />
</xsl:call-template>

And the resulting template definition:

<xsl:template name="foo">
  <xsl:param name="foo" select="$foo" />
  <xsl:param name="bar" select="$bar" />
  <xsl:param name="baz" select="$baz" />
  ...possibly more params...
</xsl:template>

This code will gives me an error “Expression error: variable ‘baz’ not found.” Is it possible to leave out the “baz” declaration?

Thank you,
Henry

Answer 1

You’re using the xsl:param syntax wrong.

Do this instead:

<xsl:template name="foo">
  <xsl:param name="foo" />
  <xsl:param name="bar" />
  <xsl:param name="baz" select="DEFAULT_VALUE" />
  ...possibly more params...
</xsl:template>

Param takes the value of the parameter passed using the xsl:with-param that matches the name of the xsl:param statement. If none is provided it takes the value of the select attribute full XPath.

More details can be found on W3School’s entry on param.