Subject: Re: [xsl] how to insert the *complete* content of a node From: "Robert P. J. Day" <rpjday@xxxxxxxxxxxxxx> Date: Fri, 21 Mar 2003 10:19:06 -0500 (EST) |
On 21 Mar 2003, danilo wrote: > Actually I do concern about attributes, so the proposed solution did not > fit at all with what I need (but thanks a lot anyway!!) > So I'm still trying with some other solution. > > Ah the: > <xsl:template match="mytag"> > <xsl:copy> > <xsl:copy-of select="*|@*"/> > </xsl:copy> > </xsl:template> > > Still does not work. > I Do not want to print the outer tag (mytag in the example) if you don't want to copy over the <mytag> tags themselves, are you saying that you want its original child elements to now be children of what used to be its parent? are you sure? try one of these: 1) <xsl:template match="mytag"> <xsl:copy> <xsl:apply-templates select="@* | node()"/> </xsl:copy> </xsl:template> 2) leave out the <xsl:copy> and </xsl:copy> as an attempt to omit the <mytag> tags themselves, but i'm leery of what your eventual tree is going to look like. rday XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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