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RE: [xsl] Print all root node attributes

Subject: RE: [xsl] Print all root node attributes
From: "Michael Kay" <mike@xxxxxxxxxxxx>
Date: Sun, 29 May 2005 21:36:37 +0100
> > <xsl:for-each select="*">
> >   <xsl:value-of select="name()" />
> >       <xsl:for-each select="@*">
> >          <xsl:value-of select="current()" />
> 
> The current() function returns the node matched by the template.
> Try
>   <xsl:value-of select="." />
> instead.
> 
> J.Pietschmann


Not so. As a self-contained XPath expression, current() returns exactly the
same as ".". It's only inside a predicate that the result is different.

Michael Kay
http://www.saxonica.com/
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