Subject: Re: [xsl] how to replace a node with a variable From: David Carlisle <davidc@xxxxxxxxx> Date: Mon, 27 Sep 2004 15:22:44 +0100 |
I need to change the following xslt so instead of the hard-coded 'customer' I can use in the pass in variable. Is there an easy way to do this? You don't have a hard coded "customer" in the code that you posted, you have "customers" and "CUSTOMER" and it isn't clear which of these you want to change, given the way you phrased yor question. The FAQ answer to selecting a node whose name is given by a parameter *say $foo) is to use *[name()=$foo] Note however that your previuous postings have suggested that no parameter is really necessary in your case as while you don't know the name of the element in your input document you know that you want to match all teh children of the document element, and for that you can just use * as discussed previously. David ________________________________________________________________________ This e-mail has been scanned for all viruses by Star. The service is powered by MessageLabs. For more information on a proactive anti-virus service working around the clock, around the globe, visit: http://www.star.net.uk ________________________________________________________________________
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
[xsl] how to replace a node with a , Paria Heidari | Thread | RE: [xsl] how to replace a node wit, Michael Kay |
[xsl] how to replace a node with a , Paria Heidari | Date | RE: [xsl] how to replace a node wit, Michael Kay |
Month |