Hi,

I have an "normal" xml,

a structure like

<x ...>
    <y ...>
      <z ...>
      </z>
     </y>
</x>

and I want to have a "flat" xml

<x .../>
<y .../>
<z .../>
so each node with its name and all attributes with original valus just not tree but with nodes not imbricated.(I have the unique ids already to know parent-child).

I try something like this:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
     version="1.0">
<xsl:output method="xml" omit-xml-declaration="yes"/>
  <xsl:template match="x | y | z">
   <xsl:text>< </xsl:text>
   <xsl:value-of select="name(.)" />
    <xsl:for-each select="@*">
    <xsl:value-of select="name()" />=<xsl:value-of select="." />
    </xsl:for-each>
    <xsl:text disable-output-escaping="yes"> /></xsl:text>
      <xsl:apply-templates select="@*|node()"/>
  </xsl:template>
</xsl:stylesheet>

but have the error on <xsl:text>< </xsl:text>.

Best regards,
white

Resolved,an root must be given.(James Durning did it)

Like this:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="2.0">
<xsl:output method="xml" omit-xml-declaration="yes"/>
<xsl:template match="/">
<ENHANCED>
<xsl:apply-templates />
</ENHANCED>
</xsl:template>
<xsl:template match="*">
<xsl:copy>
<xsl:copy-of select="@*"/>
</xsl:copy>
<xsl:apply-templates/>
</xsl:template>
</xsl:stylesheet>

Your XML is not well-formed, a < character must always be escaped as &lt;.

But your stylesheet is a bad example of tag abuse. There's no need and no good reason to use disable-output-escaping here, or to write start and end tags rather than writing nodes. In fact, I think your transformation is nothing more than

<xsl:template match="*">
  <xsl:copy>
  <xsl:copy-of select="@*"/>
  </xsl:copy>
  <xsl:apply-templates/>
</xsl:template>

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference