<xsl:value-of select="name()"/>
- Robin
Robin Delaney
FJA Feilmeier & Junker GmbH
Elsenheimerstraße 65, 80687 München.
* +49 (0) 89 - 76 901-7058
Fax +49 (0) 89 - 76 901-9502
web http://www.fja.com
* Mailto:Robin(_dot_)Delaney(_at_)fja(_dot_)com
-----Ursprüngliche Nachricht-----
Von: Simon Kelly [SMTP:kelly(_at_)ipe(_dot_)fzk(_dot_)de]
Gesendet am: Donnerstag, 20. März 2003 16:08
An: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Betreff: [xsl] Getting the value of the tag itself.
Hi all,
I know haow to get all the relevent information from a node, but I am
struggling to remeber how to get the nodes name itself. IE if I have a
node
called <table name="table1"> how do I print out it's name table, and not
the
@name table1??
Cheers
Simon
Institut fuer
Prozessdatenverarbeitung
und Elektronik,
Forschungszentrum Karlsruhe GmbH,
Postfach 3640,
D-76021 Karlsruhe,
Germany.
Tel: (+49)/7247 82-4042
E-mail : kelly(_at_)ipe(_dot_)fzk(_dot_)de
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