must be <xsl:if test="..."/>
Regards,
Joerg
katharine wykes wrote:
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(_at_)mulberrytech(_dot_)com>
Reply-To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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[(_at_)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
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