<xsl:apply-templates> effect on DOM object type?

Joseph Kesselman · Apr 12, 2007

XSLT is namespace-aware. Your stylesheet copies a <span> element with no
namespace into the output document, which means it won't be recognized
as being an XHTML span element.

Change your code to:

<xsl:template match="x">
<span xmlns="http://www.w3.org/1999/xhtml">
<svg xmlns="http://www.w3.org/2000/svg" />
</span>
</xsl:template>

C.W.Holeman II · Apr 12, 2007

I am using Firefox 2.0.0.3 on MSWindows.
I have an XML file http://emle.sourceforge.net/emle020000/testb.xml
which invokes an XSLT file http://emle.sourceforge.net/emle020000/testb.xsl

<?xml version="1.0" encoding="ISO-8859-1"?>
<?xml-stylesheet href="testb.xsl" type="text/xsl"?>
<z xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<x/>
</z>

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl

utput
method="xml"
encoding="iso-8859-1"
indent="yes"
doctype-public="-//W3C//DTD XHTML 1.0 Transitional//EN"
doctype-system="http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"
/>

<xsl:template match="/z">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script>
.... DEBUG CODE DELETED. ...
</head>
<body onload="init();">
<div>
<xsl:apply-templates select="x" />
</div>
</body>
</html>
</xsl:template>

<xsl:template match="x">
<span>
<svg xmlns="http://www.w3.org/2000/svg" />
</span>
</xsl:template>

</xsl:stylesheet>

The result of the transform is:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0
Transitional//EN""http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html
xmlns="http://www.w3.org/1999/xhtml"><head
xmlns="http://www.w3.org/1999/xhtml"><script type="text/javascript">
function init() { lu = function luf(){return
'http://www.w3.org/1999/xhtml';}; var span = document.evaluate("//span",
document, null, XPathResult.FIRST_ORDERED_NODE_TYPE, null)
..singleNodeValue; var div = document.evaluate("//xhtml:div", document,
lu, XPathResult.FIRST_ORDERED_NODE_TYPE, null) .singleNodeValue;
alert("span:" + span + " div:" + div); return; } </script></head><body
xmlns="http://www.w3.org/1999/xhtml" onload="init();"> <div> <span> <svg
xmlns="http://www.w3.org/2000/svg"/> </span> </div></body></html>The alert
displays span as an Element object and the div as an HTMLDivElement.
What is it about the <xsl:template match="x"> that has this effect?

C.W.Holeman II · Apr 12, 2007

Wrapped lines fixed.

I am using Firefox 2.0.0.3 on MSWindows.
I have an XML file http://emle.sourceforge.net/emle020000/testb.xml
which invokes an XSLT file http://emle.sourceforge.net/emle020000/testb.xsl

<?xml version="1.0" encoding="ISO-8859-1"?>
<?xml-stylesheet href="testb.xsl" type="text/xsl"?>
<z xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<x/>
</z>

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl

utput
method="xml"
encoding="iso-8859-1"
indent="yes"
doctype-public="-//W3C//DTD XHTML 1.0 Transitional//EN"
doctype-system="http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"
/>

<xsl:template match="/z">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script>
.... DEBUG CODE DELETED. ...
</head>
<body onload="init();">
<div>
<xsl:apply-templates select="x" />
</div>
</body>
</html>
</xsl:template>

<xsl:template match="x">
<span>
<svg xmlns="http://www.w3.org/2000/svg" />
</span>
</xsl:template>

</xsl:stylesheet>

The result of the transform is:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head xmlns="http://www.w3.org/1999/xhtml">
<script type="text/javascript">
function init() {
lu = function luf(){return 'http://www.w3.org/1999/xhtml';};
var span = document.evaluate("//span",
document, null, XPathResult.FIRST_ORDERED_NODE_TYPE, null)
.singleNodeValue;
var div = document.evaluate("//xhtml:div",
document, lu, XPathResult.FIRST_ORDERED_NODE_TYPE, null)
.singleNodeValue;
alert("span:" + span + " div:" + div);
return;
}
</script>
</head>
<body xmlns="http://www.w3.org/1999/xhtml" onload="init();">
<div>
<span>
<svg xmlns="http://www.w3.org/2000/svg"/>
</span>
</div>
</body>
</html>

The alert displays span as an Element object and the div as an
HTMLDivElement.
What is it about the <xsl:template match="x"> that has this effect?

Martin Honnen · Apr 12, 2007

C.W.Holeman II said:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

Add
xmlns="http://www.w3.org/1999/xhtml"
to the xsl:stylesheet element, that way all your literal XHTML elements
will be in the XHTML namespace. Otherwise in any template you write you
will need to make sure that the XHTML default namespace declaration is
set for XHTML elements.

C.W.Holeman II · Apr 12, 2007

Joseph Kesselman said:
XSLT is namespace-aware. Your stylesheet copies a <span> element with no
namespace into the output document, which means it won't be recognized as
being an XHTML span element.

Change your code to:

<xsl:template match="x">
<span xmlns="http://www.w3.org/1999/xhtml">
<svg xmlns="http://www.w3.org/2000/svg" />
</span>
</xsl:template>

Martin Honnen said:
Add
xmlns="http://www.w3.org/1999/xhtml"
to the xsl:stylesheet element, that way all your literal XHTML elements
will be in the XHTML namespace. Otherwise in any template you write you
will need to make sure that the XHTML default namespace declaration is set
for XHTML elements.

That does it. Thanks again. Corrected and working version:

http://emle.sourceforge.net/emle020000/testb2.xml
http://emle.sourceforge.net/emle020000/testb2.xsl

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<xsl:apply-templates> effect on DOM object type?

Joseph Kesselman

C.W.Holeman II

C.W.Holeman II

Martin Honnen

C.W.Holeman II

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