Subject: [xsl] sorting using a precalculated value From: Stephane Bailliez <sbailliez@xxxxxxxxxxxxxx> Date: Wed, 28 Mar 2001 15:10:51 +0100 |
I have the following XML: <classes> <class name="OuterClass1"> <class name="InnerClass"/> </class> <class name="AOuterClass"/> </classes> I want to have the following output (ie, class name must be built and the whole stuff sorted) AOuterClass OuterClass1 OuterClass1.InnerClass to build the class name I'm using: <xsl:template match="class" mode="class.name"> <xsl:if test="parent::class"> <xsl:apply-templates select="parent::class" mode="class.name"/> .<xsl:value-of select="@name"/> </xsl:if> <xsl:if test="not(parent::class)"> <xsl:value-of select="@name"/> </xsl:if> </xsl:template> However, I'm a little bit stuck here because I cannot do the following: <xsl:templates match="classes"> <xsl:for-each select=".//class"> <xsl:variable name="class.name> <xsl:apply-templates select="." mode="class.name"/> </xsl:variable> <xsl:sort select="$class.name"/> <!---- Not possible -----> <xsl:apply-templates select="." mode="class.name"/> </xsl:template> Since I need to do this sorting/name resolution many times, in different contexts, it would be nice to precalculate all this via keys, but I'm not sure I can do this. If someone has an idea how to do this, I'd be glad to know. I'm pretty sure it's simple, I'm simply missing something. Thanks a lot. -- Stéphane Bailliez Software Engineer, Paris - France iMediation - http://www.imediation.com Disclaimer: All the opinions expressed above are mine and not those from my company. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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