Re: [xsl] why this expression: match="@foo='bar'" does not work with XT ?

Subject: Re: [xsl] why this expression: match="@foo='bar'" does not work with XT ?
From: Mike Brown <mike@xxxxxxxx>
Date: Sat, 6 Jan 2001 13:50:27 -0700 (MST)
Mathieu Mangeot Lerebours wrote:
> <xsl:template match="@foo='bar'">

As Ken Holman mentioned, the solution is
<xsl:template match="@foo[.='bar']">

You also wanted to know what was wrong...

@foo='bar' is an equality test. It evaluates to a boolean
(true or false). So you are essentially saying this
template matches nodes that are true or false.

In an xsl:template's match attribute, you want a pattern:
a subset of XPath expressions that essentially identify
node-sets, not a boolean, and against which a given node
can be tested for a match.

For example if match="foo" then a given node will match if
it is an element named foo. If match="fu/foo" then the node 
will match if it is an element named foo that is a child of
an element named fu.

The main difference between a regular XPath expression and a 
pattern here is that there is no current node; instead the
processor is reading the pattern from right to left and
verifying that the current node follows the pattern.

The expression 


would normally be the nodes that are, relative to the current
node, something like current node --> child::fu --> child::foo.

In a pattern, it is as if the node being tested is the
current node, and means, essentially,
current node --> self::foo[parent::fu]

You don't really have to thoroughly get patterns. I have been
working with them for nearly 2 years and they're still hard to
explain. The important thing to understand is that an xsl:template
match pattern is not an instruction to go find nodes. That's
what xsl:apply-templates is for. The processor will have a node
already, as a result of xsl:apply-templates, usually, and will be
looking for a template that is a good match for it.

   - Mike
Mike J. Brown, software engineer at            My XML/XSL resources: in Denver, Colorado, USA    

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