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RE: problem in using * as a default parameter in xsl:param
- From: TSchutzerWeissmann at uk dot imshealth dot com
- To: xsl-list at lists dot mulberrytech dot com
- Date: Mon, 29 Apr 2002 17:17:56 +0100
- Subject: RE: [xsl] problem in using * as a default parameter in xsl:param
- Reply-to: xsl-list at lists dot mulberrytech dot com
Michael Kay wrote:
><xsl:param name="childElement" select="'*'"/>
>
><xsl:for-each
> select="element[$childElement!='*][childName=$childElement] |
> element[$childElement='*']">
>
Michael, Oleg,
This is an elegant way of making * into a wildcard, but won't it test the
text value of the children rather than their name?
If one was to use this,
select="element/*[local-name()=$childElement]", as Oleg suggests,
will * then act as a wildcard anyway?
Thanks for your help,
Tom
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