lxml question

[Uwe Schmitt]

Hi,

I have to parse some text which pretends to be XML. lxml does not want
to parse it, because it lacks a root element.
I think that this situation is not unusual, so: is there a way to
force lxml to parse it ?

My work around is wrapping the text with "<root>...</root>" before
feeding lxmls parser.

Greetings, Uwe

[Mark Thomas]

On Sep 26, 11:19*am, Uwe Schmitt <rocksportroc.. [e-mail] m>
wrote:

I have to parse some text which pretends to be XML. lxml does not want
to parse it, because it lacks a root element.
I think that this situation is not unusual, so: is there a way to
force lxml to parse it ?

By "pretends to be XML" you mean XML-like but not really XML?

My work around is wrapping the text with "<root>...</root>" before
feeding lxmls parser.

That's actually not a bad solution, if you know that the document is
otherwise well-formed. Another thing you can do is use libxml2's
"recover" mode which accommodates non-well-formed XML.

parser = etree.XMLParser (recover=True)
tree = etree.XML(your_ xml_string, parser)

You'll still need to use your wrapper root element, because recover
mode will ignore everything after the first root closes (and it won't
throw an error).

-- Mark.

[alex23]

On Sep 27, 1:19*am, Uwe Schmitt <rocksportroc.. [e-mail] m>
wrote:

I have to parse some text which pretends to be XML. lxml does not want
to parse it, because it lacks a root element.

Another option is BeautifulSoup, which handles badly formed XML really
well:

http://www.crummy.com/software/BeautifulSoup/

[Stefan Behnel]

Uwe Schmitt wrote:

I have to parse some text which pretends to be XML. lxml does not want
to parse it, because it lacks a root element.
I think that this situation is not unusual, so: is there a way to
force lxml to parse it ?

My work around is wrapping the text with "<root>...</root>" before
feeding lxmls parser.

Yes, you can do that. To avoid creating an intermediate string, you can use
the feed parser and do something like this:

parser = etree.XMLParser ()
parser.feed("<r oot>")
parser.feed(you r_xml_tag_seque nce_data)
parser.feed("</root>")
root = parser.close()

Stefan