Re: build-in template question
Garvin Riensche wrote, On 2008-09-05 00:17:Hello,During evaluation of applicable templates on a particular node, the most specific template is chosen. If two or more templates is of the same level of specification, then the template with the highest priority is chosen. I have forgotten what happens if two templates cannot be separated (a situaiton that I try to avoid), although a qualified guess is that an XSLT 1.0 processor does something and an XSLT 2.0 processor returns an error message. Anyway, in this case, the match="e" is more specifc than match="text()|@*" so the first template is chosen.
I think the "match='e'" doesn't compete with the "match='text()|@*'" because "e" is neither a text, nor an attribute node. So that template for e is chosen which copies the element node "e". In the definition of xsl:copy it is said:
"If the context item is a node, evaluating the xsl:copy instruction constructs a copy of the context node, and the result of the xsl:copy instruction is this newly constructed node. By default, the namespace nodes of the context node are automatically copied as well, but the attributes and children of the node are not automatically copied."
Because "text" is a child of "e" I think it is not copied by xsl:copy.
But If "e" if replaced by the sequence constructor of the template how can the build-in template match the text-node, which got already replaced?
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