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Re: [xsl] deep copy without attributes
Subject: Re: [xsl] deep copy without attributes From: Graydon <graydon@xxxxxxxxx> Date: Mon, 20 May 2013 19:02:44 -0400 |
On Mon, May 20, 2013 at 10:34:31PM +0100, Ihe Onwuka scripsit: > On Mon, May 20, 2013 at 8:48 PM, Graydon <graydon@xxxxxxxxx> wrote: > > You can't make the solution simpler than the problem. > > But you can make the solution more complicated than the problem. > > Say I wanted to copy a directory structure but not carry over the > symbolic links. > > I'd expect to be able to accomplish that by setting an option on my > copy command rather than having to implement a recursive shell script. Symbolic links, in XML terms, would be elements from some other tree inserted by reference. XML doesn't do that, and if it did do that, there'd be a case for some kind of "is it really there, or referenced?" distinction. (Attributes, by file system analogy, would be things like mtime and ctime.) What you were describing, if I'm remembering it right, is wanting a result tree that was just like your source tree only without the attribute nodes. <xsl:template match="node()"> <xsl:copy> <xsl:apply-templates select="node()"/> </xsl:copy> </xsl:template> does that; it's not recursive, it's a single template. It'll get applied to all the nodes, but that's the magic of the XSLT declarative model. -- Graydon Saunders graydon@xxxxxxxxx
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