Subject: Re: [xsl] Problem with doctype-system in transformation xml to xml From: "Axel Kesten" <kesten@xxxxxxxxxxxxxxx> Date: Wed, 7 Dec 2005 17:23:33 +0100 |
Hi David thank you for the hint. I've done as you mentioned (not exactly) but now i have another problem. Now i have all the attributes of the xsl:stylesheet element as attributes of my top-level element in the destination xml-file. The stylesheet now looks like this: <xsl:template match="/"> <xsl:choose> <xsl:when test="/C/D/E[position()=1]"> <A> <xsl:for-each select="/C/D/E"> <xsl:variable name="doc_name" select="."></xsl:variable> <xsl:apply-templates select="document($doc_name)"></xsl:apply-templates> </xsl:for-each> </A> </xsl:when> <xsl:otherwise> <xsl:variable name="nodename" select="A/CATEGORY"></xsl:variable> <xsl:choose> <xsl:when test="$nodename='test'"> <xsl:call-template name="copy"></xsl:call-template> </xsl:when> </xsl:choose> </xsl:otherwise> </xsl:choose> </xsl:template> <xsl:template name="Kopie"> <xsl:copy-of select="MSRSW/SW-SYSTEMS/SW-SYSTEM/SW-COMPONENT-SPEC/*"></xsl:copy-of> </xsl:template> </xsl:stylesheet> The elements stylesheet and output i have not touched, except of the doctype-system. Is it a problem that i have my top-level element <A> not as first statement in the root template? Kind regards Axel ----- Original Message ----- From: "David Carlisle" <davidc@xxxxxxxxx> To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx> Sent: Wednesday, December 07, 2005 2:31 PM Subject: Re: [xsl] Problem with doctype-system in transformation xml to xml > > Because of the structure of the xml-files i'm evaluating, the </A> is > written to the output tree after the <A>. > > that is your problem, you have not generated an A element just written > some text that looks like <A> so as far as XSLT is concerned your top > level generated element is B.
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