Subject: Re: [xsl] xsl:copy- of and namespaces From: "Alessandro Bologna" <alessandro.bologna@xxxxxxxxx> Date: Tue, 1 Apr 2008 12:35:03 -0400 |
Off topic. This list will never cease to amaze me. 3 valid answers, with explanations, in 9 minutes. Kudos to all. Alessandro On Tue, Apr 1, 2008 at 11:19 AM, Florent Georges <lists@xxxxxxxxxxxx> wrote: > Garvin Riensche wrote: > > Hi > > > > Can someone please tell me how I can give elements of a > > subtree that are copied to the output with <xsl:copy-of > > select="."> a new namespace? > > The aim of xsl:copy-of is to, well, copy its arguments. > If I understand you right (but that's not sure), you want to > change the namespace URI part of the name of the elements to > copy. So that's not excatly a copy, but a transformation: > > <xsl:template match="@*|node()"> > <xsl:copy> > <xsl:apply-templates select="@*|node()"/> > </xsl:copy> > </xsl:template> > > <xsl:template match="*" priority="0"> > <xsl:element name="new:{ local-name(.) }"> > <xsl:apply-templates select="@*|node()"/> > </xsl:element> > </xsl:template> > > That means: copy everything, except the element nodes, in this case > create a new element node whose the name is the same local-name but in > the new namespace. > > Regards, > > --drkm > > > > > > > > > > > > > > > > > > > > > > > > _____________________________________________________________________________ > Envoyez avec Yahoo! Mail. Plus de moyens pour rester en contact. http://mail.yahoo.fr
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] xsl:copy- of and namespac, Florent Georges | Thread | [xsl] removing unecessary attribute, Andreas Peter |
Re: [xsl] xsl:copy- of and namespac, Florent Georges | Date | Re: [xsl] xsl:copy- of and namespac, Garvin Riensche |
Month |