[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] 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
|
PURCHASE STYLUS STUDIO ONLINE TODAY!Purchasing Stylus Studio from our online shop is Easy, Secure and Value Priced! Download The World's Best XML IDE!Accelerate XML development with our award-winning XML IDE - Download a free trial today! Subscribe in XML format
|