RE: does xsl:output method xml guarentee well formedness?

In the presence of character maps or disable-output-escaping, there is no
guarantee that the serialized result of a transformation will be
well-formed.

This is stated at the end of the introduction to section 5 of the
serialization spec, 

http://www.w3.org/TR/xslt-xquery-serialization/#xml-output

Michael Kay
http://www.saxonica.com/

-----Original Message-----
From: James Fuller [mailto:jim(_dot_)fuller(_at_)ruminate(_dot_)co(_dot_)uk] 
Sent: 11 April 2005 09:38
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] does xsl:output method xml guarentee well formedness?

for example...in XSLT 2.0 (SAXON 8.4)

ex. XML
<?xml version="1.0" encoding="UTF-8"?>
<example>
    <test>a</test>
</example>

ex. XSLT
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="2.0">
    <xsl:output method="xml" indent="yes" use-character-maps="test"/>
    <xsl:template match="example">
        <xsl:value-of select="test"/>
    </xsl:template>
    <xsl:character-map name="test">
        <xsl:output-character character="a" string="&lt;test&gt;"/>
    </xsl:character-map>
</xsl:stylesheet>

results in

<?xml version="1.0" encoding="UTF-8"?><test>

using the following stylesheet as an analogy;

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="2.0">
    <xsl:output method="xml" indent="yes" />
    <xsl:template match="example">
        &lt;test&gt;
    </xsl:template>
</xsl:stylesheet>

results in

<?xml version="1.0" encoding="UTF-8"?>

        &lt;test&gt;

which is of course well formed.

cant really tell from the spec if xsl:output method guarentee's well
formedness or perhaps xsl:character-map needs something more ?

--Jim Fuller


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