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Re: [xsl] Using attributes with XPath


Subject: Re: [xsl] Using attributes with XPath
From: David Carlisle <davidc@xxxxxxxxx>
Date: Wed, 23 Aug 2006 23:35:29 +0100

<xsl:template match="list[parent::description]">
you could write that as match="description/list" it means the same, but
looks nicer (you can just use match="list" unless you have some other
template for other list elements.
<fo:list-block>
  <xsl:apply-templates/>
that is the same as   <xsl:apply-templates select="node()"/> so only
applies nodesto child attributes, you want to apply nodes to attributes
as well, which would be  <xsl:apply-templates select="@*|node()"/>
 </fo:list-block>
</xsl:template>


<xsl:template match="@type[parent::list]">
again you could write that as list/@type
<fo:inline  font-size="10pt" font-weight="normal"> 
<xsl:apply-templates/>   
This applies templates to the child nodes of ths node, but attributes
don't have children so it will always produce nothing. Perhaps you meant
<xsl:value-of select="."/>
  </fo:inline>
</xsl:template>

but this would then make an fo:inline as a child of fo:list-block, which
I don't think is allowed in FO (XSLT won't mind, it'll just do what you
ask)

David


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