It seems to me that you are going about this the wrong way. The XSLT document()
function produces an in-memory tree representation of the XML document. There
are no "tags". Instead of using the XSLT, try instead to include it as an
ordinary text file.
--
Charles Knell
cknell(_at_)onebox(_dot_)com - email
-----Original Message-----
From: <Lisa(_dot_)Bergqvist(_at_)hm(_dot_)com>
Sent: Fri, 21 Apr 2006 15:11:51 +0200
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Subject: [xsl] SV: xsl-list Digest 21 Apr 2006 05:10:00 -0000 Issue 753
I am writing an XSL:FO document that will create a pdf document. One
chapter will contain an "example XML".
I access the XML file by a variable using document() function, and so
far so good. I now want to list the whole file tags, names, values and
all, pereferrably also indentaions.
When I try to list the whole XML file firstly, I try:
<xsl:for-each select="$SourceMetadata/*">
<!--List the XML file-->
<xsl:copy-of select="." />
</xsl:for-each>
This results in an error message saying: "Element name :
BusinessObjectName.(namespace = Null) Source context
unavailable.Unsupported element encountered".
The BusinessObjectName is the first element in the XML file.
The XML file does not contain any namespace.
I then try to add attribute copy-namespaces=No, then the parser says it
is wrong attribute for copy-of. Parser = Apache FOP.bat.
I change the code to:
<xsl:for-each select="$SourceMetadata/*">
<!--List the XML file-->
<xsl:copy-of select="node()" />
</xsl:for-each>
Now I get all element values in a row. No element names and no tags.
I then change to:
<xsl:for-each select="$SourceMetadata/*">
<!--List the XML file-->
<xsl:copy-of select="name()/>
</xsl:for-each>
I then get all tag names in a row, no values, no tags.
What can I do?
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