Re: [xsl] forwarding only existing parameter

[David Carlisle <davidc@xxxxxxxxx>]

> 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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