Re: why this expression: match="@foo='bar'" does not w

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 

  fu/foo

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: 
webb.net in Denver, Colorado, USA              http://skew.org/xml/


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