Re: [xsl] stop the processing of default attributes?

[David Carlisle <davidc@xxxxxxxxx>]

On 19/08/2011 01:45, Dan Vint wrote:
> I am trying to get my output to replicate the input as best possible.

then you want to replace

<xsl:element  name="{name()}">
                <xsl:for-each select="@*">
                        <xsl:attribute name="{name(.)}"><xsl:value-of 
select="."/></xsl:attribute>
                </xsl:for-each>

by

<xsl:copy>
  <xsl:copy-of select="@*"/>

the namespace differences are that for both xsl:element and 
xsl:attribute, given an input of x:foo then
xsl:copy(-of)
will generate a node with the same namespace and local name as the 
source 9and if possible the same prefix)

whereas

<xsl:element  name="{name()}">

will work like a literal element in the stylesheet:

<x:foo>

which will generate an element with prefix x and local name foo and 
whatever namespace URI is bound to x _in the stylesheet_ the namespace 
of x:foo in the source will not be used.


David


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