J
Jeff.M
Consider the following XSL:
<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/
Transform" xmlns
hp="http://php.net/xsl">
<xslutput method="html"
doctype-public="-//W3C//DTD HTML 4.01//EN"
doctype-system="http://www.w3.org/TR/html4/strict.dtd" />
<xsl:template match="/">
<html>
<head>
<title>Hello</title>
</head>
<body>
<p>fubar</p>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
The presence of the xmlnshp="http://php.net/xsl" XML attribute
causes the same to appear as an attribute of the HTML element in my
output, like so:
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/
TR/html4/strict.dtd">
<html xmlnshp="http://php.net/xsl">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Hello</title>
</head>
<body><p>fubar</p></body>
</html>
The problem is this isn't valid HTML. Without sacrificing using PHP
functions as XSLT functions, how do I get this attribute out of my
output?
<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/
Transform" xmlns
hp="http://php.net/xsl"><xsl
utput method="html"doctype-public="-//W3C//DTD HTML 4.01//EN"
doctype-system="http://www.w3.org/TR/html4/strict.dtd" />
<xsl:template match="/">
<html>
<head>
<title>Hello</title>
</head>
<body>
<p>fubar</p>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
The presence of the xmlns
hp="http://php.net/xsl" XML attributecauses the same to appear as an attribute of the HTML element in my
output, like so:
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/
TR/html4/strict.dtd">
<html xmlns
hp="http://php.net/xsl"><head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Hello</title>
</head>
<body><p>fubar</p></body>
</html>
The problem is this isn't valid HTML. Without sacrificing using PHP
functions as XSLT functions, how do I get this attribute out of my
output?