Subject: Re: [xsl] Node selection based on parent attribute From: Joerg Heinicke <joerg.heinicke@xxxxxx> Date: Fri, 30 Aug 2002 19:48:25 +0200 |
Still trying to get a better understanding of xslt. Using the code you specified below brings up a parser error:
Attribute 'select' is invalid on 'xsl:if'.
Any clues as to the cause of this? Cheers,
From: Wendell Piez <wapiez@xxxxxxxxxxxxxxxx> Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: [xsl] Node selection based on parent attribute Date: Fri, 30 Aug 2002 12:24:31 -0400
Katharine,
Another approach (besides the ones using for-each) is just to use the built-in descent of the tree, as in something like:
<xsl:template match="menu"> <xsl:if select="ancestor::menu[@id=$id']"> <xsl:value-of select="@id"/> <!-- copies out this menu's @id if it has an ancestor menu with @id = $id --> </xsl:if> <xsl:apply-templates/> <!-- continues the tree traversal in case there are any below --> </xsl:template>
Make sure the parameter is set to the *value* of the id ('1', '6', whatever) whose descendants you want.
Cheers, Wendell
System Development VIRBUS AG Fon +49(0)341-979-7419 Fax +49(0)341-979-7409 joerg.heinicke@xxxxxxxxx www.virbus.de
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