The ElementTree.parse reads from a file, how can I use this if I already have the XML data in a string?
Maybe I am missing something here, but there must be a way to use the ElementTree without writing out the string to a file and reading it again.
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4 Answers
You can parse the text as a string, which creates an Element, and create an ElementTree using that Element.
import xml.etree.ElementTree as ET
tree = ET.ElementTree(ET.fromstring(xmlstring))
I just came across this issue and the documentation, while complete, is not very straightforward on the difference in usage between the parse() and fromstring() methods.
3 Comments
Anton Tarasenko
The second line can be simply
root = ET.fromstring(xmlstring). Equals ET.parse('file.xml').getroot(): docs.python.org/3.6/library/…
batbrat
@Anton, as the OP states, the idea is to generate an ElementTree, and not an Element. This is useful, for instance, when you want to use ElementTree.write().
Louie Bafford
If anyone is curious this also works for
lxmlIf you're using xml.etree.ElementTree.parse to parse from a file, then you can use xml.etree.ElementTree.fromstring to get the root Element of the document. Often you don't actually need an ElementTree.
8 Comments
Samuel Lampa
The problem is that ElementTree.fromstring generates an element, and not an ElementTree! Anyone knows how to work around this?
Siddharth Menon
Same problem as @SamuelLampa mentioned. I is not a ElementTree, I am not able to do
getroot() for this
Colin Pickard
@SamuelLampa see dgassaway's answer, use
ET.ElementTree(ET.fromstring(xmlstring))
2.718
for the correct answer, see the one provided by @dgassaway
Stevoisiak
Don't forget the import statements
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You need the xml.etree.ElementTree.fromstring(text)
from xml.etree.ElementTree import XML, fromstring
myxml = fromstring(text)
1 Comment
rjurney
Yes, but this is not an
ElementTree it is the root Element. This API is different and can't do all the same things.io.StringIO is another option for getting XML into xml.etree.ElementTree:
import io
f = io.StringIO(xmlstring)
tree = ET.parse(f)
root = tree.getroot()
Hovever, it does not affect the XML declaration one would assume to be in tree (although that's needed for ElementTree.write()). See How to write XML declaration using xml.etree.ElementTree.
