Burak Arslan, 09.05.2014 18:52:
> On 05/09/14 16:55, Stefan Behnel wrote:
>> ElementTree has gained a nice API in
>> Py3.4 that supports this in a much saner way than SAX, using iterators.
>> Basically, you just dump in some data that you received and get back an
>> iterator over the elements (and their subtrees) that it generated from it.
>> Intercept on the right top elements and you get your next subtree as soon
>> as it's ready.
>
> Here's a small script:
A bit hard to read, though.
> events = etree.iterparse(istr, events=("start", "end"))
> stack = deque()
> for event, element in events:
> if event == "start":
> stack.append(element)
> elif event == "end":
> stack.pop()
>
> if len(stack) == 0:
> break
>
> print(istr.tell(), "%5s, %4s, %s" % (event, element.tag, element.text))
>
> where istr is an input-stream. (Fully working example:
> https://gist.github.com/plq/025005a71e8135c46800)
>
> I was expecting to have istr.tell() return the position where the first
> root element ends, which would make it possible to continue parsing with
> another call to etree.iterparse(). But istr.tell() returns the position
> of EOF after the first call to next() on the iterator it returns.
Correct, because it finished parsing. It controls the reading process and
reads ahead, that's how iterparse() works.
> Without the stack check, the loop eventually throws an exception and the
> offset value in that exception is None.
>
> So I'm lost here, how it'd possible to parse OP's document with lxml?
See my earlier post. Instead of XMLParser, just use the XMLPullParser for
incremental (non-blocking) parsing and processing.
To make this clear, though: to use an XML parser, you need well formed XML,
and that means exactly one root element.
Stefan
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