Question :
I want to use the method of “findall” to locate some elements of the source xml file in the ElementTree module.
However, the source xml file (test.xml) has namespace. I truncate part of xml file as sample:
<?xml version="1.0" encoding="iso-8859-1"?>
<XML_HEADER xmlns="http://www.test.com">
<TYPE>Updates</TYPE>
<DATE>9/26/2012 10:30:34 AM</DATE>
<COPYRIGHT_NOTICE>All Rights Reserved.</COPYRIGHT_NOTICE>
<LICENSE>newlicense.htm</LICENSE>
<DEAL_LEVEL>
<PAID_OFF>N</PAID_OFF>
</DEAL_LEVEL>
</XML_HEADER>
The sample python code is below:
from xml.etree import ElementTree as ET
tree = ET.parse(r"test.xml")
el1 = tree.findall("DEAL_LEVEL/PAID_OFF") # Return None
el2 = tree.findall("{http://www.test.com}DEAL_LEVEL/{http://www.test.com}PAID_OFF") # Return <Element '{http://www.test.com}DEAL_LEVEL/PAID_OFF' at 0xb78b90>
Although it can works, because there is a namespace “{http://www.test.com}”, it’s very inconvenient to add a namespace in front of each tag.
How can I ignore the namespace when using the method of “find”, “findall” and so on?
Answer #1:
Instead of modifying the XML document itself, it’s best to parse it and then modify the tags in the result. This way you can handle multiple namespaces and namespace aliases:
from io import StringIO # for Python 2 import from StringIO instead
import xml.etree.ElementTree as ET
# instead of ET.fromstring(xml)
it = ET.iterparse(StringIO(xml))
for _, el in it:
prefix, has_namespace, postfix = el.tag.partition('}')
if has_namespace:
el.tag = postfix # strip all namespaces
root = it.root
This is based on the discussion here:
http://bugs.python.org/issue18304
Update: rpartition
instead of partition
makes sure you get the tag name in postfix
even if there is no namespace. Thus you could condense it:
for _, el in it:
_, _, el.tag = el.tag.rpartition('}') # strip ns
Answer #2:
If you remove the xmlns attribute from the xml before parsing it then there won’t be a namespace prepended to each tag in the tree.
import re
xmlstring = re.sub(' ]+"', '', xmlstring, count=1)
Answer #3:
The answers so far explicitely put the namespace value in the script. For a more generic solution, I would rather extract the namespace from the xml:
import re
def get_namespace(element):
m = re.match('{.*}', element.tag)
return m.group(0) if m else ''
And use it in find method:
namespace = get_namespace(tree.getroot())
print tree.find('./{0}parent/{0}version'.format(namespace)).text
Answer #4:
Here’s an extension to nonagon’s answer, which also strips namespaces off attributes:
from StringIO import StringIO
import xml.etree.ElementTree as ET
# instead of ET.fromstring(xml)
it = ET.iterparse(StringIO(xml))
for _, el in it:
if '}' in el.tag:
el.tag = el.tag.split('}', 1)[1] # strip all namespaces
for at in list(el.attrib.keys()): # strip namespaces of attributes too
if '}' in at:
newat = at.split('}', 1)[1]
el.attrib[newat] = el.attrib[at]
del el.attrib[at]
root = it.root
UPDATE: added list()
so the iterator works (needed for Python 3)
Answer #5:
Improving on the answer by ericspod:
Instead of changing the parse mode globally we can wrap this in an object supporting the with construct.
from xml.parsers import expat
class DisableXmlNamespaces:
def __enter__(self):
self.oldcreate = expat.ParserCreate
expat.ParserCreate = lambda encoding, sep: self.oldcreate(encoding, None)
def __exit__(self, type, value, traceback):
expat.ParserCreate = self.oldcreate
This can then be used as follows
import xml.etree.ElementTree as ET
with DisableXmlNamespaces():
tree = ET.parse("test.xml")
The beauty of this way is that it does not change any behaviour for unrelated code outside the with block. I ended up creating this after getting errors in unrelated libraries after using the version by ericspod which also happened to use expat.
Answer #6:
You can use the elegant string formatting construct as well:
ns='http://www.test.com'
el2 = tree.findall("{%s}DEAL_LEVEL/{%s}PAID_OFF" %(ns,ns))
or, if you’re sure that PAID_OFF only appears in one level in tree:
el2 = tree.findall(".//{%s}PAID_OFF" % ns)
Answer #7:
If you’re using ElementTree
and not cElementTree
you can force Expat to ignore namespace processing by replacing ParserCreate()
:
from xml.parsers import expat
oldcreate = expat.ParserCreate
expat.ParserCreate = lambda encoding, sep: oldcreate(encoding, None)
ElementTree
tries to use Expat by calling ParserCreate()
but provides no option to not provide a namespace separator string, the above code will cause it to be ignore but be warned this could break other things.
Answer #8:
I might be late for this but I dont think re.sub
is a good solution.
However the rewrite xml.parsers.expat
does not work for Python 3.x versions,
The main culprit is the xml/etree/ElementTree.py
see bottom of the source code
# Import the C accelerators
try:
# Element is going to be shadowed by the C implementation. We need to keep
# the Python version of it accessible for some "creative" by external code
# (see tests)
_Element_Py = Element
# Element, SubElement, ParseError, TreeBuilder, XMLParser
from _elementtree import *
except ImportError:
pass
Which is kinda sad.
The solution is to get rid of it first.
import _elementtree
try:
del _elementtree.XMLParser
except AttributeError:
# in case deleted twice
pass
else:
from xml.parsers import expat # NOQA: F811
oldcreate = expat.ParserCreate
expat.ParserCreate = lambda encoding, sep: oldcreate(encoding, None)
Tested on Python 3.6.
Try try
statement is useful in case somewhere in your code you reload or import a module twice you get some strange errors like
- maximum recursion depth exceeded
- AttributeError: XMLParser
btw damn the etree source code looks really messy.