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NotXMLError: Failed to parse the XML data

I'm trying to use the Entrez module from Biopython to retrive full text articles from PubMed Central. This is my code to do the same.
import urllib3
import json
import requests
from Bio import Entrez
from Bio.Entrez import efetch, Parser
print(Parser.__file__)
pmcid = 'PMC2837563'
def print_text(pmcid):
handle = efetch(db='pmc', id=pmcid, retmode='xml', rettype=None)
#print(handle.read())
record = Entrez.read(handle)
print(record)
print_text(pmcid)
handle.read() works which means the data is being fetched properly. But, I'm not able to do Entrez.read(handle) to convert the fetched data into a python object. It gives me the below error:
NotXMLError: Failed to parse the XML data (syntax error: line 1036, column 69). Please make sure that the input data are in XML format.
Could someone tell me what to do about this? This seems to be correct syntax as per the biopython documentation.

The reason is that the last available Biopython version (1.79) does not recognise DTD with uri http://www.niso.org/schemas/ali/1.0/. The GitHub version has the corrected Parser but it is not available from pip now.
Compare:
current 1.79
def startNamespaceDeclHandler(self, prefix, uri):
"""Handle start of an XML namespace declaration."""
if prefix == "xsi":
# This is an xml schema
self.schema_namespace = uri
self.parser.StartElementHandler = self.schemaHandler
else:
# Note that the DTD for MathML specifies a default attribute
# that declares the namespace for each MathML element. This means
# that MathML element in the XML has an invisible MathML namespace
# declaration that triggers a call to startNamespaceDeclHandler
# and endNamespaceDeclHandler. Therefore we need to count how often
# startNamespaceDeclHandler and endNamespaceDeclHandler were called
# to find out their first and last invocation for each namespace.
if prefix == "mml":
assert uri == "http://www.w3.org/1998/Math/MathML"
elif prefix == "xlink":
assert uri == "http://www.w3.org/1999/xlink"
else:
raise ValueError("Unknown prefix '%s' with uri '%s'" % (prefix, uri))
self.namespace_level[prefix] += 1
self.namespace_prefix[uri] = prefix
GitHub
def startNamespaceDeclHandler(self, prefix, uri):
"""Handle start of an XML namespace declaration."""
if prefix == "xsi":
# This is an xml schema
self.schema_namespace = uri
self.parser.StartElementHandler = self.schemaHandler
else:
# Note that the DTD for MathML specifies a default attribute
# that declares the namespace for each MathML element. This means
# that MathML element in the XML has an invisible MathML namespace
# declaration that triggers a call to startNamespaceDeclHandler
# and endNamespaceDeclHandler. Therefore we need to count how often
# startNamespaceDeclHandler and endNamespaceDeclHandler were called
# to find out their first and last invocation for each namespace.
if prefix == "mml":
assert uri == "http://www.w3.org/1998/Math/MathML"
elif prefix == "xlink":
assert uri == "http://www.w3.org/1999/xlink"
elif prefix == "ali":
assert uri == "http://www.niso.org/schemas/ali/1.0/"
else:
raise ValueError(f"Unknown prefix '{prefix}' with uri '{uri}'")
self.namespace_level[prefix] += 1
self.namespace_prefix[uri] = prefix
So you can either exchange or edit Parser.py file, or use third party libraries for converting your handle to built-in python object.
If you want download just a full text of the article, you could try to download a pdf through metapub & go on to extract a text via textract.
import metapub
from urllib.request import urlretrieve
import textract
pmcid = 'PMC2837563'
fetch = metapub.PubMedFetcher()
article_metadata = fetch.article_by_pmcid(pmcid)
#Get just an abstract
abstract = article_metadata.abstract
#Download full article text
pmid = article_metadata.pmid
url = metapub.FindIt(pmid).url
urlretrieve(url, any_path)
with open(another_path, "w") as textfile:
textfile.write(textract.process(
any_path,
extension='pdf',
method='pdftotext',
encoding="utf_8",
))

Building a generic XML parser in Python?

I am a newbie and having 1 week experience writing python scripts.
I am trying to write a generic parser (Library for all my future jobs) which parses any input XML without any prior knowledge of tags.
Parse input XML.
Get the values from the XML and Set the values basing on the tags.
Use these values in the rest of the job.
I am using the "xml.etree.ElementTree" library and i am able to parse the XML in the below mentioned way.
#!/usr/bin/python
import os
import xml.etree.ElementTree as etree
import logging
logging.basicConfig(level=logging.DEBUG)
logger = logging.getLogger(__name__)
logger.info('start reading XML property file')
filename = "mood_ib_history_parameters_DEV.xml"
logger.info('getting the current location')
__currentlocation__ = os.getcwd()
__fullpath__ = os.path.join(__currentlocation__,filename)
logger.info('start parsing the XML property file')
tree = etree.parse(__fullpath__)
root = tree.getroot()
hive_db = root.find("hive_db").text
EDGE_HIVE_CONN = root.find("EDGE_HIVE_CONN").text
target_dir = root.find("target_dir").text
to_email_alias = root.find("to_email_alias").text
to_email_cc = root.find("to_email_cc").text
from_email_alias = root.find("from_email_alias").text
dburl = root.find("dburl").text
SQOOP_EDGE_CONN = root.find("SQOOP_EDGE_CONN").text
user_name = root.find("user_name").text
password = root.find("password").text
IB_log_table = root.find("IB_log_table").text
SR_DG_master_table = root.find("SR_DG_master_table").text
SR_DG_table = root.find("SR_DG_table").text
logger.info('Hive DB %s', hive_db)
logger.info('Hive DB %s', hive_db)
logger.info('Edge Hive Connection %s', EDGE_HIVE_CONN)
logger.info('Target Directory %s', target_dir)
logger.info('To Email address %s', to_email_alias)
logger.info('CC Email address %s', to_email_cc)
logger.info('From Email address %s', from_email_alias)
logger.info('DB URL %s',dburl)
logger.info('Sqoop Edge node connection %s',SQOOP_EDGE_CONN)
logger.info('Log table name %s',IB_log_table)
logger.info('Master table name %s',SR_DG_master_table)
logger.info('Data governance table name %s',SR_DG_table)
Now the question is if i want to parse an XML without any knowledge of the tags and elements and use the values how do i do it. I have gone through multiple tutorials but all of them help me with parsing the XML by using the tags like below
SQOOP_EDGE_CONN = root.find("SQOOP_EDGE_CONN").text
Can anybody point me to a right tutorial or library or a code snippet to parse the XML dynamically.

I think official documentation is pretty clear and contains some examples: https://docs.python.org/3/library/xml.etree.elementtree.html
The main part you need to implement is loop over the child nodes (potentially recursively):
for child in root:
# child.tag contains the tag name, child.attrib contains the attributes
print(child.tag, child.attrib)

Well parsing is easy as that - etree.parse(path)
Once you've got the root in hand using tree.getroot() you can just iterate over the tree using Python's "in":
for child_node in tree.getroot():
print child_node.text
Then, to see tags these child_nodes have, you do the same trick.
This lets you go over all tags in the XML without having to know the tag names at all.

Error validating/parsing xml file against xsd with lxml/objectify in Python

in Python/Django, I need to parse and objectify a file .xml according to a given XMLSchema made of three .xsd files referring each other in such a way:
schema3.xsd (referring schema1.xsd)
schema2.xsd (referring schema1.xsd)
schema1.xsd (referring schema2.xsd)
xml schemas import
For this I'm using the following piece of code which I've already tested being succesfull when used with a couple of xml/xsd files (where .xsd is "standalone" without refering others .xsd):
import lxml
import os.path
from lxml import etree, objectify
from lxml.etree import XMLSyntaxError
def xml_validator(request):
# define path of files
path_file_xml = '../myxmlfile.xml'
path_file_xsd = '../schema3.xsd'
# get file XML
xml_file = open(path_file_xml, 'r')
xml_string = xml_file.read()
xml_file.close()
# get XML Schema
doc = etree.parse(path_file_xsd)
schema = etree.XMLSchema(doc)
#define parser
parser = objectify.makeparser(schema=schema)
# trasform XML file
root = objectify.fromstring(xml_string, parser)
test1 = root.tag
return render(request, 'risultati.html', {'risultato': test1})
Unfortunately, I'm stucked with the following error that i got with the multiple .xsd described above:
complex type 'ObjectType': The content model is not determinist.
Request Method: GET Request URL: http://127.0.0.1:8000/xml_validator
Django Version: 1.9.1 Exception Type: XMLSchemaParseError Exception
Value: complex type 'ObjectType': The content model is not
determinist., line 80
Any idea about that ?
Thanks a lot in advance for any suggestion or useful tips to approach this problem...
cheers
Update 23/03/2016
Here (and in the following answers to the post, because it actually exceed the max number of characters for a post), a sample of the files to figure out the problem...
sample files on GitHub

My best guess would be that your XSD model does not obey the Unique Particle Attribution rule. I would rule that out before looking at anything else.

Python XML validation with SAX, pyexpat, ElementTree

I'm trying to check validity of XML files (against DTDs, entities, Processing instructions, namespaces) in Python 3.4.
Looking at Python docs the default underlying parser for three Python XML modules pyexpat, ELementTree and SAX is expat. On Pyexpat page (https://docs.python.org/3.4/library/pyexpat.html?highlight=pyexpat#module-xml.parsers.expat) says that non-validating version of expat parser is used:
"The xml.parsers.expat module is a Python interface to the Expat non-validating XML parser." Yet at the same time when you look into SAX documentation in Python you see all these handler functions for enabling DTD validation etc. How the heck do you make them work?
However, according to this post Parsing XML Entity with python xml.sax SAX can validate. Obviously with expat as a parser.
I have reused the code from this post and can't get it to work I get error saying expat does not support validation:
"File "/usr/lib/python3.4/xml/sax/expatreader.py", line 149, in setFeature
"expat does not support validation")
xml.sax._exceptions.SAXNotSupportedException: expat does not support validation".
In the post Python 2.5 was used, so maybe SAX has changed since then...
This is the code:
import xml.sax
from xml.sax import handler, make_parser, parse
import os
import collections
class SaxParser():
# initializer with directory part as argument
def __init__(self, dir_path):
self.dir_path = dir_path
def test_each_file(self, file_path):
# ensure full file name is shown
rev = file_path[::-1] # reverse string file_path to access position of "/"
file = file_path[-rev.index("/"):]
try:
f = open(file_path, 'r', encoding="ISO-8859-1") # same as "latin-1" encoding
# see this for enabling validation:
# https://stackoverflow.com/questions/6349513/parsing-xml-entity-with-python-xml-sax
parser = make_parser() # default parser is expat
parser.setContentHandler(handler.ContentHandler())
parser.setFeature(handler.feature_namespaces,True)
parser.setFeature(handler.feature_validation,True)
parser.setFeature(handler.feature_external_ges, True)
parser.parse(f)
f.close()
return (file, "OK")
except xml.sax.SAXParseException as PE:
column = PE.getColumnNumber()
line = PE.getLineNumber()
msg = PE.getMessage()
value = msg + " " + str(line) + " " + str(column)
return (file, value)
except ValueError:
return (file, "ValueError. DTD uri not found.") # that can happen
def test_directory_sax(self, dir_path):
tuples = []
for ind, file in enumerate(os.listdir(dir_path), 1):
if file.endswith('.xml'):
tuples.append(self.test_each_file(dir_path + file))
# convert into dict and sort it by key (file number)
dict_of_errors = dict(tuples)
dict_of_errors = collections.OrderedDict(sorted(dict_of_errors.items()))
return dict_of_errors
# ========================================================================
# INVOKE TESTS FOR SINGLE SPECIFIED DIRECTORY THAT CONTAINS TEST FILES
# ========================================================================
path = # path to directory where xml file is. - not the filepath!
single_sax = SaxParser(path)
print('============================================================')
print('TEST FOR SAX parser FOR DIRECTORY ' + path)
print('============================================================\n')
print(single_sax.test_directory_sax(path))
and test xml file (should produce validation error):
<!DOCTYPE root [
<!ATTLIST root
id2 ID "x23"
>
]>
<!-- an ID attribute must have a declared default
of #IMPLIED or #REQUIRED
-->
<root/>
How do I check validity? For either one of three XML modules?
A simple example would do.
Thanks.

If you look into the source file, you'll see that the xml.sax.handler.feature_validation is not really doing anything but raising this exception:
def setFeature(self, name, state):
# ...
elif name == feature_validation:
if state:
raise SAXNotSupportedException(
"expat does not support validation")
#...
I would suggest using lxml to do this. An example would be like this:
from lxml import etree
from cStringIO import StringIO
# from io import StringIO (py3)
f = StringIO('<!ATTLIST root id2 ID "x23">')
dtd = etree.DTD(f)
root = etree.XML('<root/>')
print(dtd.validate(root))
print(dtd.error_log.filter_from_errors()[0])

lxml, xi:include, and original file

I'm using lxml to parse a file that contains xi:include elements, and I'm resolve the includes using xinclude().
Given an element, is there any way to identify the file and source line that the element originally appeared in?
For example:
from lxml import etree
doc = etree.parse('file.xml')
doc.xinclude()
xpath_expression = ...
elt = doc.xpath(xpath_expression)
# Print file name and source line of `elt` location

The xinclude expansion will add an xml:base attribute to the top level expanded element,
and elt.base and elt.sourceline are also updated for the child nodes as well, so:
print elt.base, elt.sourceline
will give you what you want.
If elt is not part of the xinclude expansion, then elt.base will point to the base
document ( 'file.xml' ) and elt.sourceline will be the line number in that file.
( Note that sourceline usually seems to actually point to the line where the element tag
ends, not to the line where it begins, if the element is on multiple lines, just as
validation error messages usually point to the closing tag where the error occurs. )
You can find the initial xincluded elements and check this with:
xels = doc.xpath( '//*[#xml:base] )
for x in xels:
print x.tag, x.base, x.sourceline
for c in x.getchildren():
print c.tag, c.base, c.sourceline

Sadly, current versions of lxml no longer include this ability. However, I've developed a workaround using a simple custom loader. Here's a test script which demonstrates the bug in the approach above along with the workaround. Note that this approach only updates the xml:base attribute of the root tag of the included document.
The output of the program (using Python 3.9.1, lxml 4.6.3):
Included file was source.xml; xinclude reports it as document.xml
Included file was source.xml; workaround reports it as source.xml
Here's the sample program.
# Includes
# ========
from pathlib import Path
from textwrap import dedent
from lxml import etree as ElementTree
from lxml import ElementInclude
# Setup
# =====
# Create a sample document, taken from the `Python stdlib
# <https://docs.python.org/3/library/xml.etree.elementtree.html#id3>`_...
Path("document.xml").write_text(
dedent(
"""\
<?xml version="1.0"?>
<document xmlns:xi="http://www.w3.org/2001/XInclude">
<xi:include href="source.xml" parse="xml" />
</document>
"""
)
)
# ...and the associated include file.
Path("source.xml").write_text("<para>This is a paragraph.</para>")
# Failing xinclude case
# =====================
# Load and xinclude this.
tree = ElementTree.parse("document.xml")
tree.xinclude()
# Show that the ``base`` attribute refers to the top-level
# ``document.xml``, instead of the xincluded ``source.xml``.
root = tree.getroot()
print(f"Included file was source.xml; xinclude reports it as {root[0].base}")
# Workaround
# ==========
# As a workaround, define a loader which sets the ``xml:base`` of an
# xincluded element. While lxml evidently used to do this, a change
# eliminated this ability per some `discussion
# <https://mail.gnome.org/archives/xml/2014-April/msg00015.html>`_,
# which included a rejected patch fixing this problem. `Current source
# <https://github.com/GNOME/libxml2/blob/master/xinclude.c#L1689>`_
# lacks this patch.
def my_loader(href, parse, encoding=None, parser=None):
ret = ElementInclude._lxml_default_loader(href, parse, encoding, parser)
ret.attrib["{http://www.w3.org/XML/1998/namespace}base"] = href
return ret
new_tree = ElementTree.parse("document.xml")
ElementInclude.include(new_tree, loader=my_loader)
new_root = new_tree.getroot()
print(f"Included file was source.xml; workaround reports it as {new_root[0].base}")