Subject: Re: [xsl] XPath question From: Jeni Tennison <mail@xxxxxxxxxxxxxxxx> Date: Fri, 24 Aug 2001 11:00:16 +0100 |
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 |
---|
|
<- Previous | Index | Next -> |
---|---|---|
[xsl] XPath question, Adam Van Den Hoven | Thread | [xsl] Announce: XSLT Cookbook, Paul Prescod |
Re: [xsl] number("Infinity")?, Jeni Tennison | Date | [xsl] convert xml file to flat file, Gurnandan Kaur |
Month |