xsl-list
[Top] [All Lists]

[xsl] Selecting all elements which have an attribute that matches a given pattern

2010-08-18 02:58:02

Hi all and apologies if this has been dealt with previously, I searched
the archives and FAQ but did not find anything that was close to my
problem:

I have an XML file (a schema in fact) that is many thousands of lines
long and I need to edit it (i.e transform it).  There are numerous
elements of type <xs:element> all of which have a "name" attribute.
Some of those name attributes end with the string "List", e.g.
"MilkTypeList".  For those elements, I want to apply a template to add
an additional <xs:element>.  I am having great trouble crafting the
correct "match" for my <xsl:template>.  I have tried contains() and
matches() but my editor (Oxygen) states that those cannot be used as a
pattern and my reading supports that.  Can anyone help?

Thanks

Stephen



Please help Logica to respect the environment by not printing this email  / 
Pour contribuer comme Logica au respect de l'environnement, merci de ne pas 
imprimer ce mail /  Bitte drucken Sie diese Nachricht nicht aus und helfen Sie 
so Logica dabei, die Umwelt zu schützen. /  Por favor ajude a Logica a 
respeitar o ambiente nao imprimindo este correio electronico.



This e-mail and any attachment is for authorised use by the intended 
recipient(s) only. It may contain proprietary material, confidential 
information and/or be subject to legal privilege. It should not be copied, 
disclosed to, retained or used by, any other party. If you are not an intended 
recipient then please promptly delete this e-mail and any attachment and all 
copies and inform the sender. Thank you.



--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--

<Prev in Thread] Current Thread [Next in Thread>