Schimon Jehudah via lxml - The Python XML Toolkit schrieb am 12.01.25 um 08:53:
On Fri, 10 Jan 2025 17:28:00 +0100
jholg--- via lxml - The Python XML Toolkit <lxml@python.org> wrote:
>>> from lxml import etree
>>> elem = etree.fromstring('<data/>')
>>> tree = elem.getroottree()
>>> tree.getroot().addprevious(etree.ProcessingInstruction('xml-stylesheet',
>>> 'type="text/xml" href="whatever.xsl"'))
>>> etree.tostring(tree)
b'<?xml-stylesheet type="text/xml" href="whatever.xsl"?><data/>'
>>>
This is almost what I need.
Sample
------
<?xml version="1.0"?><feed xmlns="http://www.w3.org/2005/Atom";></feed>
Result
------
>>> etree.tostring(tree)
b'<?xml-stylesheet type="text/xml" href="whatever.xsl"?><feed
xmlns="http://www.w3.org/2005/Atom";>\n</feed>'
Node root is missing. XML is not valid.
It is well-formed XML. The XML declaration ("<?xml …>") is not mandatory
for UTF-8 encoded XML.
("valid" means that it adheres to a schema. There is no XML validation
involved here.)
Desired result
--------------
b'<?xml version="1.0"?><?xml-stylesheet type="text/xml" href="whatever.xsl"?><feed
xmlns="http://www.w3.org/2005/Atom";>\n</feed>'
Node stylesheet appears after node root.
I suppose, that I would have to copy node root and concatenate it to
the, node stylesheet instruction.
Just configure the output encoding explicitly or force the XML declaration
to be written.
https://lxml.de/tutorial.html#serialisation
Stefan
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