Subject: Re: [xsl] variable question From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> Date: Wed, 10 Nov 2004 22:00:19 +0000 |
Hi Bruce, I haven't been following this thread, just saw my name mentioned... Wendell said: >> Maybe Mike or Jeni or someone can suggest an easier way to do this in >> XSLT 2.0. There isn't any way of dynamically evaluating strings as XPaths in XSLT 2.0 -- you still have to generate the stylesheet to do it. But of course, if you're using Saxon you can use the saxon:evaluate() extension function. > Maybe use a key? > > I'm confused (again!). I'm basically trying to rework some of my > code, in part based around Geert's suggestions, but I can't really > see how to do what I'm wanting to do (which is to be able to work on > content from external files). It seems, for example, that I cannot > use a key on content I want to access via the doc function. > > <xsl:variable name="bibrecord" select="doc(concat('bib-data/', > $bibkey, '.mods'))" /> > <xsl:key name="biblio" match="$bibrecord//mods:mods" use="@ID" /> The document that's searched when you call the key() function is the current document at the time of the call. So you should do: <xsl:key name="biblio" match="mods:mods" use="@ID" /> and then something along the lines of: <xsl:for-each select="$bibrecord/key('biblio', $bibkey)"> ... </xsl:for-each> though I think that you need $bibkey to be defined as: <xsl:variable name="bibkey" select="//db:bilioref/@linkend" /> since the <biblioref> elements can occur anywhere within the (source) document. Note that the path "$bibrecord/key('biblio', $bibkey)" is newly allowed in XPath 2.0, and makes searching documents using keys a whole lot easier than it used to be. Cheers, Jeni --- Jeni Tennison http://www.jenitennison.com/
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