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Re: forwarding only existing parameter
- To: xsl-list at lists dot mulberrytech dot com
- Subject: Re: [xsl] forwarding only existing parameter
- From: David Carlisle <davidc at nag dot co dot uk>
- Date: Mon, 17 Sep 2001 12:46:48 +0100
- References: <20010917102614.4F75D4893@silbermann.snv.jussieu.fr>
- Reply-To: xsl-list at lists dot mulberrytech dot com
> I'm trying to forward parameters from a template to another, only if they are
> defined, in the most simple way.
variables and parameters are always defined in XSLT. If a definition of
a variable x is not in scope then you can not use $x so there is no
possibilty of asking if x is defined.
Why can't you just always pass on the value of your parameter?
by choosing a suitable default value you should be able to determine
which case you are in in your code.
David
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