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#1
July 22nd, 2003, 01:54 PM
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XSLT outputing <?xml> and <!DOCTYPE>
I'm using XSLT to transform XML data into XHTML, but I need to output the xml declaration and the doctype before the <html> tag...
<xsl:template match="/"> <?xml version="1.0"?> <!DOCTYPE html PUBLIC "-//OPENWAVE//DTD XHTML Mobile 1.0//EN" "http://www.openwave.com/DTD/xhtml-mobile10.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> ... </xsl:template> How can I have these included in my code? Currently they cause errors.
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#2
July 22nd, 2003, 02:11 PM
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i believe it has to go in the xsl:output element which must appear as the child node of the xsl:stylesheet element. I hope this helps.
Code:
<xsl:output method="xml" encoding="iso-8859-1" doctype-system="http://www.w3.org/TR/2000/REC-xhtml1-20000126/DTD/xhtml1-strict.dtd"; doctype-pubilc="-//W3C//DTD XHTML 1.0 Strict//EN"/>
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#3
July 22nd, 2003, 02:32 PM
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So it is not possible to have the following outputted by my XSL? The above example would probably take care of the doctype, but what about the <?xml version="1.0"?>
<?xml version="1.0"?> <!DOCTYPE html PUBLIC "-//OPENWAVE//DTD XHTML Mobile 1.0//EN" "http://www.openwave.com/DTD/xhtml-mobile10.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> ... </html>
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#4
July 22nd, 2003, 02:39 PM
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you could out put it but i am not sure that it will not give unexpected results. typically the "processor directive" is the first line in your xsl page. something like what you have. It is basically a statement that says "hello I ma a xml file... process me!" Then the stylesheets requires everything to be placed inside the <xsl:stylesheet> element for transformation.
Code:
<?xml version="1.0" encoding="UTF-8"?>
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#5
July 30th, 2003, 12:33 AM
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This is controlled by the "omit-xml-declaration" attribute, which is generally set by "no" by default. If your output method is "xml" then it should automatically output the declaration. If it doesn't, try adding omit-xml-declaration="no" to the <xsl:output> element.
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