Subject: [xsl] Position() of parent node From: Simon Cansick <SC@xxxxxxxxxxxxxxxxxxx> Date: Tue, 6 Feb 2001 18:09:33 -0000 |
Can anyone provide me with the syntax for getting the position() value of the current nodes' parent node (and the parent parent etc. position() value). I seem only able to return the current position(). I need to use the value in my generated HTML, not for template selection. Thanks for any help. Simon Cansick sc@xxxxxxxxxxxxxxxxxxx Access Accounting Ltd --------------------------------------------------------------- The Old School, Stratford St Mary, Colchester, Essex. CO7 6LZ, United Kingdom. Phone +44 (0) 1206 322575 Fax +44 (0) 1206 322956 Internet http://www.access-accounts.com http://www.accessweb.co.uk --------------------------------------------------------------- Access Accounts - The Brand Leader in e-commerce financials This message is confidential; its contents do not constitute a commitment by Access Accounting Ltd except where provided for in a written agreement between you and Access Accounting Ltd. Any opinions expressed within this email are those of the author and are not necessarily endorsed by Access Accounting Ltd. Any unauthorised disclosure, use or dissemination, either whole or partial, is prohibited. If you are not the intended recipient of the message, please notify the sender immediately. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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