[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message]

Re: XPath question

Subject: Re: XPath question
From: Jeni Tennison <mail@xxxxxxxxxxxxxxxx>
Date: Fri, 24 Aug 2001 11:00:16 +0100
Re:  XPath question
Hi Adam,

> Now what is all the all the attrib nodes from $rules whose name AND parents
> state attribute are not present together in $attribs.

I don't think it's possible to do in a single XPath. You can do:

  <xsl:for-each select="$rules/attrib">
    <xsl:if test="not($attribs[@name = current()/@name and
                               @state = current()/../@state])">
      ...
    </xsl:if>
  </xsl:for-each>

If the attrib elements in $attribs were held somewhere other than in a
variable, then you could construct a key for them:

<xsl:key name="attribs" match="attrib"
         use="concat(@name, ' ', @state)" />

And then you could do:

  $rules/attrib[not(key('attribs', concat(@name, ' ', ../@state)))]

I hope that helps,

Jeni

---
Jeni Tennison
http://www.jenitennison.com/


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list


Current Thread

PURCHASE STYLUS STUDIO ONLINE TODAY!

Purchasing Stylus Studio from our online shop is Easy, Secure and Value Priced!

Buy Stylus Studio Now

Download The World's Best XML IDE!

Accelerate XML development with our award-winning XML IDE - Download a free trial today!

Don't miss another message! Subscribe to this list today.
Email
First Name
Last Name
Company
Subscribe in XML format
RSS 2.0
Atom 0.3
Site Map | Privacy Policy | Terms of Use | Trademarks
Free Stylus Studio XML Training:
W3C Member
Stylus Studio® and DataDirect XQuery ™are products from DataDirect Technologies, is a registered trademark of Progress Software Corporation, in the U.S. and other countries. © 2004-2013 All Rights Reserved.