i have a xml file.i trying to read it in a usual way as shown below
def xmlfilereadread(self,path):
doc = minidom.parse(path)
Account = doc.getElementsByTagName("sf:ReceiverSet")[0]
num = Account.getAttribute('totalNo')
aList = []
for i in range(int(num)):
print(i)
AccountReference = doc.getElementsByTagName("sf:Receiver")[i]
but i need to use panda unstead of this code.how can i read data.my sample xml code is
<?xml version="1.0" encoding="UTF-8"?>
<sf:IFile xmlns:sf="http://www.canadapost.ca/smartflow" sequenceNo="10">
<sf:ReceiverSet documentTypes="TAXBILL" organization="lincolntax" totalNo="3">
<sf:Receiver sequenceNo="1" correlationID="1114567890123456789">
<sf:AccountReference>11145678901234567891111</sf:AccountReference>
<sf:SubscriptionAuth> <sf:ParamSet>
<sf:Param name="auth1">1114567890123456789</sf:Param>
<sf:Param name="auth2">CARTER, JOE</sf:Param> </sf:ParamSet>
</sf:SubscriptionAuth>
</sf:Receiver> <sf:Receiver sequenceNo="2" correlationID="2224567890123456789">
<sf:AccountReference>22245678901234567892222</sf:AccountReference> <sf:SubscriptionAuth> <sf:ParamSet>
<sf:Param name="auth1">2224567890123456789</sf:Param>
<sf:Param name="auth2">DOE, JANE</sf:Param> </sf:ParamSet>
</sf:SubscriptionAuth> </sf:Receiver> <sf:Receiver sequenceNo="3" correlationID="3334567890123456789">
<sf:AccountReference>33345678901234567893333</sf:AccountReference> <sf:SubscriptionAuth> <sf:ParamSet>
<sf:Param name="auth1">3334567890123456789</sf:Param> <sf:Param name="auth2">SOZE, KEYSER</sf:Param>
</sf:ParamSet> </sf:SubscriptionAuth> </sf:Receiver> </sf:ReceiverSet> </sf:IFile>
XML is an inherently hierarchical data format, and the most natural
way to represent it is with a tree. ET has two classes for this
purpose - ElementTree represents the whole XML document as a tree, and
Element represents a single node in this tree. Interactions with the
whole document (reading and writing to/from files) are usually done on
the ElementTree level. Interactions with a single XML element and its
sub-elements are done on the Element level
.
import xml.etree.ElementTree as ET
tree = ET.parse('country_data.xml')
root = tree.getroot()
Or you can use lxml
from lxml import etree
root = etree.parse(r'local-path-to-.xml')
print (etree.tostring(root))
Related
I am trying to parse out all the green highlighted attributes (some sensitive things have been blacked out), I have a bunch of XML files all with similar formats, I already know how to loop through all of them individually them I am having trouble parsing out the specific attributes though.
XML Document
I need the text in the attributes: name="text1"
from
project logLevel="verbose" version="2.0" mainModule="Main" name="text1">
destinationDir="/text2" from
put label="Put Files" destinationDir="/Trigger/FPDMMT_INBOUND">
destDir="/text3" from
copy disabled="false" version="1.0" label="Archive Files" destDir="/text3" suffix="">
I am using
import csv
import os
import re
import xml.etree.ElementTree as ET
tree = ET.parse(XMLfile_path)
item = tree.getroot()[0]
root = tree.getroot()
print (item.get("name"))
print (root.get("name"))
This outputs:
Main
text1
The item.get pulls the line at index [0] which is the first line root in the tree which is <module
The root.get pulls from the first line <project
I know there's a way to search for exactly the right part of the root/tree with something like:
test = root.find('./project/module/ftp/put')
print (test.get("destinationDir"))
I need to be able to jump directly to the thing I need and output the attributes I need.
Any help would be appreciated
Thanks.
Simplified copy of your XML:
xml = '''<project logLevel="verbose" version="2.0" mainModule="Main" name="hidden">
<module name="Main">
<createWorkspace version="1.0"/>
<ftp version="1.0" label="FTP connection to PRD">
<put label="Put Files" destinationDir="destination1">
</put>
</ftp>
<ftp version="1.0" label="FTP connection to PRD">
<put label="Put Files" destinationDir="destination2">
</put>
</ftp>
<copy disabled="false" destDir="destination3">
</copy>
</module>
</project>
'''
# solution using ETree
from xml.etree import ElementTree as ET
root = ET.fromstring(xml)
name = root.get('name')
ftp_destination_dir1 = root.findall('./module/ftp/put')[0].get('destinationDir')
ftp_destination_dir2 = root.findall('./module/ftp/put')[1].get('destinationDir')
copy_destination_dir = root.find('./module/copy').get('destDir')
print(name)
print(ftp_destination_dir1)
print(ftp_destination_dir2)
print(copy_destination_dir)
# solution using lxml
from lxml import etree as et
root = et.fromstring(xml)
name = root.get('name')
ftp_destination_dirs = root.xpath('./module/ftp/put/#destinationDir')
copy_destination_dir = root.xpath('./module/copy/#destDir')[0]
print(name)
print(ftp_destination_dirs[0])
print(ftp_destination_dirs[1])
print(copy_destination_dir)
Would you help me, pleace, to get an access to elemnt with name 'id' by the following construction in Python (i have lxml and xml.etree.ElementTree libraries).
Desirable result: '0000000'
Desirable method:
Search in xml-document a child, where it's name is fcsProtocolEF3.
Search in fcsProtocolEF3 an element with name 'id'.
It is crucial to search by element name. Not by ordinal position.
I tried to use something like this: tree.findall('{http://zakupki.gov.ru/oos/export/1}fcsProtocolEF3')[0].findall('{http://zakupki.gov.ru/oos/types/1}id')[0].text
it works, but it requires to input namespaces. XML-document have different namespaces and I don't know how to define them beforehand.
Thank you.
That would be great to use something like XQuery in SQL:
value('(/*:export/*:fcsProtocolEF3/*:id)[1]', 'nvarchar(21)')) AS [id],
XML-document:
<?xml version="1.0" encoding="UTF-8" standalone="true"?>
<ns2:export xmlns:ns3="http://zakupki.gov.ru/oos/common/1" xmlns:ns4="http://zakupki.gov.ru/oos/base/1" xmlns:ns2="http://zakupki.gov.ru/oos/export/1" xmlns:ns10="http://zakupki.gov.ru/oos/printform/1" xmlns:ns11="http://zakupki.gov.ru/oos/control99/1" xmlns:ns9="http://zakupki.gov.ru/oos/SMTypes/1" xmlns:ns7="http://zakupki.gov.ru/oos/pprf615types/1" xmlns:ns8="http://zakupki.gov.ru/oos/EPtypes/1" xmlns:ns5="http://zakupki.gov.ru/oos/TPtypes/1" xmlns:ns6="http://zakupki.gov.ru/oos/CPtypes/1" xmlns="http://zakupki.gov.ru/oos/types/1">
<ns2:fcsProtocolEF3 schemeVersion="10.2">
<id>0000000</id>
<purchaseNumber>0000000000000000</purchaseNumber>
</ns2:fcsProtocolEF3>
</ns2:export>
lxml solution:
xml = '''<?xml version="1.0"?>
<ns2:export xmlns:ns3="http://zakupki.gov.ru/oos/common/1" xmlns:ns4="http://zakupki.gov.ru/oos/base/1" xmlns:ns2="http://zakupki.gov.ru/oos/export/1" xmlns:ns10="http://zakupki.gov.ru/oos/printform/1" xmlns:ns11="http://zakupki.gov.ru/oos/control99/1" xmlns:ns9="http://zakupki.gov.ru/oos/SMTypes/1" xmlns:ns7="http://zakupki.gov.ru/oos/pprf615types/1" xmlns:ns8="http://zakupki.gov.ru/oos/EPtypes/1" xmlns:ns5="http://zakupki.gov.ru/oos/TPtypes/1" xmlns:ns6="http://zakupki.gov.ru/oos/CPtypes/1" xmlns="http://zakupki.gov.ru/oos/types/1">
<ns2:fcsProtocolEF3 schemeVersion="10.2">
<id>0000000</id>
<purchaseNumber>0000000000000000</purchaseNumber>
</ns2:fcsProtocolEF3>
</ns2:export>'''
from lxml import etree as et
root = et.fromstring(xml)
text = root.xpath('//*[local-name()="export"]/*[local-name()="fcsProtocolEF3"]/*[local-name()="id"]/text()')[0]
print(text)
Below is ET based solution. NS are in use.
import xml.etree.ElementTree as ET
xml = '''<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<ns2:export xmlns:ns3="http://zakupki.gov.ru/oos/common/1" xmlns:ns4="http://zakupki.gov.ru/oos/base/1" xmlns:ns2="http://zakupki.gov.ru/oos/export/1" xmlns:ns10="http://zakupki.gov.ru/oos/printform/1" xmlns:ns11="http://zakupki.gov.ru/oos/control99/1" xmlns:ns9="http://zakupki.gov.ru/oos/SMTypes/1" xmlns:ns7="http://zakupki.gov.ru/oos/pprf615types/1" xmlns:ns8="http://zakupki.gov.ru/oos/EPtypes/1" xmlns:ns5="http://zakupki.gov.ru/oos/TPtypes/1" xmlns:ns6="http://zakupki.gov.ru/oos/CPtypes/1" xmlns="http://zakupki.gov.ru/oos/types/1">
<ns2:fcsProtocolEF3 schemeVersion="10.2">
<id>0000000</id>
<purchaseNumber>0000000000000000</purchaseNumber>
</ns2:fcsProtocolEF3>
</ns2:export>
'''
def get_id_text():
root = ET.fromstring(xml)
fcs = root.find('{http://zakupki.gov.ru/oos/export/1}fcsProtocolEF3')
# assuming there is one fcs element and one id under fcs
return fcs.find('{http://zakupki.gov.ru/oos/types/1}id').text
print(get_id_text())
output
0000000
I am having problems generating a XML document using the ElementTree framework in Python 3. I tried registering the namespace before setting up the document. Right now it seems that I can generate a XML document only by adding the namespace to each element like a=Element("{full_namespace_URI}element_name") which seems tedious.
How do I setup the default namespace and can omit putting it in each element?
Any help is appreciated.
I have written a small demo program for Python 3:
from io import BytesIO
from xml.etree import ElementTree as ET
ET.register_namespace("", "urn:dslforum-org:service-1-0")
"""
desired output
==============
<?xml version='1.0' encoding='utf-8'?>
<topNode xmlns="urn:dslforum-org:service-1-0"">
<childNode>content</childNode>
</topNode>
"""
# build XML document without namespaces
a = ET.Element("topNode")
b = ET.Element("childNode")
b.text = "content"
a.append(b)
tree = ET.ElementTree(a)
# build XML document with namespaces
a_ns = ET.Element("{dsl}topNode")
b_ns = ET.Element("{dsl}childNode")
b_ns.text = "content"
a_ns.append(b_ns)
tree_ns = ET.ElementTree(a_ns)
def print_element_tree(element_tree, comment, default_namespace=None):
"""
print element tree with comment to standard out
"""
with BytesIO() as buf:
element_tree.write(buf, encoding="utf-8", xml_declaration=True,
default_namespace=default_namespace)
buf.seek(0)
print(comment)
print(buf.read().decode("utf-8"))
print_element_tree(tree, "Element Tree without XML namespace")
print_element_tree(tree_ns, "Element Tree with XML namespace", "dsl")
I believe you are overthinking this.
Registering a default namespace in your code avoids the ns0: aliases.
Registering any namespaces you will use while creating a document allows you to designate the alias used for each namespace.
To achieve your desired output, assign the namespace to your top element:
a = ET.Element("{urn:dslforum-org:service-1-0}topNode")
The preceding ET.register_namespace("", "urn:dslforum-org:service-1-0") will make that the default namespace in the document, assign it to topNode, and not prefix your tag names.
<?xml version='1.0' encoding='utf-8'?>
<topNode xmlns="urn:dslforum-org:service-1-0"><childNode>content</childNode></topNode>
If you remove the register_namespace() call, then you get this monstrosity:
<?xml version='1.0' encoding='utf-8'?>
<ns0:topNode xmlns:ns0="urn:dslforum-org:service-1-0"><childNode>content</childNode></ns0:topNode>
I have this XML
<Body>
<Batch_Number>2000</Batch_Number>
<Total_No_Of_Batches>12312</Total_No_Of_Batches>
<requestNo>1923</requestNo>
<Parent1>
<Parent2>
<Parent3>
<lastModifiedDateTime>2022-11-11T11:07:30.000</lastModifiedDateTime>
<purpose>NeverMore</purpose>
<endDate>9999-12-31T00:00:00.000</endDate>
<createdDateTime>2019-06-06T06:32:16.000</createdDateTime>
<createdOn>2019-06-06T08:32:16.000</createdOn>
<address2>Forever street 21</address2>
<externalCode>code123</externalCode>
<lastModifiedBy>user2.thisUser</lastModifiedBy>
<lastModifiedOn>2039-06-11T13:07:30.000</lastModifiedOn>
<lastModifiedBy>MG</lastModifiedBy>
<PS>1234431</PS>
</Parent3>
</Parent2>
</Parent1>
</Body>
Is there a way to return the value for lastModifiedBy for example where the path has this specific structure :
Body.Parent1.Parent2.Parent3.lastModifiedBy
Idealy, I would like to populate a dictionary with the child tag name and its value, for example :
dict[lastModifiedBy.tag] = lastModifiedBy.text
You can use xml from standart library for working with xml files.
from xml.etree import ElementTree as ET
tree = ET.parse("d.xml") # our xml file
root = tree.getroot()
And then you can access elements as indexes or you can use root like as a list:
for i in root:
print(i)
A XML element may have more than one child with same tag name (even you have two lastModifiedBy in the Parent3). This is why we use them like lists, they works like a list. So you shouldn't try to use them like dictionary.
I think you need to use XPath. Like so:
from xml.etree import ElementTree as ET
tree = ET.parse("d.xml") # our xml file
root = tree.getroot()
s = root.findall(".Parent1/Parent2/Parent3/lastModifiedBy")
for i in s:
print(i.text)
This gives you all lastModifiedBy elements in the Parent3 element. You can access to any index if you want too, like this:
from xml.etree import ElementTree as ET
tree = ET.parse("d.xml") # our xml file
root = tree.getroot()
s = root.find(".Parent1/Parent2/Parent3/lastModifiedBy[1]") # first element with "lastModifiedBy" tag
print(s.text)
I'm trying to write a simple program to read my financial XML files from GNUCash, and learn Python in the process.
The XML looks like this:
<?xml version="1.0" encoding="utf-8" ?>
<gnc-v2
xmlns:gnc="http://www.gnucash.org/XML/gnc"
xmlns:act="http://www.gnucash.org/XML/act"
xmlns:book="http://www.gnucash.org/XML/book"
{...}
xmlns:vendor="http://www.gnucash.org/XML/vendor">
<gnc:count-data cd:type="book">1</gnc:count-data>
<gnc:book version="2.0.0">
<book:id type="guid">91314601aa6afd17727c44657419974a</book:id>
<gnc:count-data cd:type="account">80</gnc:count-data>
<gnc:count-data cd:type="transaction">826</gnc:count-data>
<gnc:count-data cd:type="budget">1</gnc:count-data>
<gnc:commodity version="2.0.0">
<cmdty:space>ISO4217</cmdty:space>
<cmdty:id>BRL</cmdty:id>
<cmdty:get_quotes/>
<cmdty:quote_source>currency</cmdty:quote_source>
<cmdty:quote_tz/>
</gnc:commodity>
Right now, i'm able to iterate and get results using
import xml.etree.ElementTree as ET
r = ET.parse("file.xml").findall('.//')
after manually cleaning the namespaces, but I'm looking for a solution that could either read the entries regardless of their namespaces OR remove the namespaces before parsing.
Note that I'm a complete noob in python, and I've read: Python and GnuCash: Extract data from GnuCash files, Cleaning an XML file in Python before parsing and python: xml.etree.ElementTree, removing "namespaces" along with ElementTree docs and I'm still lost...
I've come up with this solution:
def strip_namespaces(self, tree):
nspOpen = re.compile("<\w*:", re.IGNORECASE)
nspClose = re.compile("<\/\w*:", re.IGNORECASE)
for i in tree:
start = re.sub(nspOpen, '<', tree.tag)
end = re.sub(nspOpen, '<\/', tree.tag)
# pprint(finaltree)
return
But I'm failing to apply it. I can't seem to be able to retrieve the tag names as they appear on the file.
I think below python code will be helpfull to you.
sample.xml
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<gnc:prodinfo xmlns:gnc="http://www.gnucash.org/XML/gnc"
xmlns:act="http://www.gnucash.org/XML/act"
xmlns:book="http://www.gnucash.org/XML/book"
xmlns:vendor="http://www.gnucash.org/XML/vendor">
<gnc:change>
<gnc:lastUpdate>2018-12-21
</gnc:lastUpdate>
</gnc:change>
<gnc:bill>
<gnc:billAccountNumber>1234</gnc:billAccountNumber>
<gnc:roles>
<gnc:id>111111</gnc:id>
<gnc:pos>2</gnc:pos>
<gnc:genid>15</gnc:genid>
</gnc:roles>
</gnc:bill>
<gnc:prodtyp>sales and service</gnc:prodtyp>
</gnc:prodinfo>
PYTHON CODE: to remove xmlns for root tag.
import xml.etree.cElementTree as ET
def xmlns(str):
str1 = str.split('{')
l=[]
for i in str1:
if '}' in i:
l.append(i.split('}')[1])
else:
l.append(i)
var = ''.join(l)
return var
tree=ET.parse('sample.xml')
root = tree.getroot()
print(root.tag) #returns root tag with xmlns as prefix
print(xmlns(root.tag)) #returns root tag with out xmlns as prefix
Output:
{http://www.gnucash.org/XML/gnc}prodinfo
prodinfo