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how to parse XML with namespace and attribute in Python?

hi I am trying to parse xml with namespace and attribute.
I am almost close by using root.findall() and .get()
However still struggling to get the accurate values from xml file.
How to get the xml attribute values ?
Input:
<?xml version="1.0" encoding="UTF-8"?><message:GenericData
xmlns:message="http://www.sdmx.org/resources/sdmxml/schemas/v2_1/message"
xmlns:common="http://www.sdmx.org/resources/sdmxml/schemas/v2_1/common"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:generic="http://www.sdmx.org/resources/sdmxml/schemas/v2_1/data/generic"
xsi:schemaLocation="http://www.sdmx.org/resources/sdmxml/schemas/v2_1/message https://sdw-
wsrest.ecb.europa.eu:443/vocabulary/sdmx/2_1/SDMXMessage.xsd
http://www.sdmx.org/resources/sdmxml/schemas/v2_1/common https://sdw-
wsrest.ecb.europa.eu:443/vocabulary/sdmx/2_1/SDMXCommon.xsd
http://www.sdmx.org/resources/sdmxml/schemas/v2_1/data/generic https://sdw-
wsrest.ecb.europa.eu:443/vocabulary/sdmx/2_1/SDMXDataGeneric.xsd">
<generic:Obs>
<generic:ObsDimension value="1999-01"/>
<generic:ObsValue value="0.7029125"/>
</generic:Obs>
<generic:Obs>
<generic:ObsDimension value="1999-02"/>
<generic:ObsValue value="0.688505"/>
</generic:Obs>
Code:
import xml.etree.ElementTree as ET
tree = ET.parse("file.xml")
root = tree.getroot()
for x in root.findall('.//'):
print(x.tag, " ", x.get('value'))
Output:
{http://www.sdmx.org/resources/sdmxml/schemas/v2_1/data/generic}Obs None
{http://www.sdmx.org/resources/sdmxml/schemas/v2_1/data/generic}ObsDimension 1999-01
{http://www.sdmx.org/resources/sdmxml/schemas/v2_1/data/generic}ObsValue 0.7029125
{http://www.sdmx.org/resources/sdmxml/schemas/v2_1/data/generic}Obs None
{http://www.sdmx.org/resources/sdmxml/schemas/v2_1/data/generic}ObsDimension 1999-02
{http://www.sdmx.org/resources/sdmxml/schemas/v2_1/data/generic}ObsValue 0.688505
Expected_Output:
1999-01 0.7029125
1999-02 0.688505

How about this:
for parent in root:
print(' '.join([child.get('value', "") for child in parent]))

Inserting an existing root into an existing Python ElementTree

I'm trying to link two existing Python ElementTree objects together.
import xml.etree.ElementTree as ET
root = ET.Element('Hello')
root2 = ET.Element('World')
node = ET.SubElement(root2, 'country')
node.text = 'Belgium'
When printed
print(ET.tostring(root))
print(ET.tostring(root2))
I get
b'<Hello />'
b'<World><country>Belgium</country></World>'
How do I add root2 to root, to get the result? `
print(ET.tostring(root))
b'<Hello><World><country>Belgium</country></World></Hello>'

How about
import xml.etree.ElementTree as ET
hello = ET.Element('Hello')
world = ET.Element('World')
hello.insert(0,world)
country = ET.SubElement(world,'Country')
country.text = 'Belgium'
print(ET.tostring(hello))
Output
b'<Hello><World><Country>Belgium</Country></World></Hello>'

It seems, that I can use the same syntax as in lists
root.append(root2)
print(ET.tostring(root))
b'<Hello><World><country>Belgium</country></World></Hello>'

Reg adding data to an existing XML in Python

I have to parse an xml file & modify the data in a particular tag using Python. I'm using Element Tree to do this. I'm able to parse & reach the required tag. But I'm not able to modify the value. I'm not sure if Element Tree is okay or if I should use TreeBuilder for this.
As you can see below I just want to replace the Not Executed under Verdict with a string value.
-<Procedure>
<PreCondition>PRECONDITION: - ECU in extended diagnostic session (zz = 0x03) </PreCondition>
<PostCondition/>
<ProcedureID>428495</ProcedureID>
<SequenceNumber>2</SequenceNumber>
<CID>-1</CID>
<**Verdict** Writable="true">NotExecuted</Verdict>
</Procedure>
import xml.etree.ElementTree as etree
X_tree = etree.parse('DIAGNOSTIC SERVER.xml')
X_root = X_tree.getroot()
ATC_Name = X_root.iterfind('TestOrder//TestOrder//TestSuite//')
try:
while(1):
temp = ATC_Name.next()
if temp.tag == 'ProcedureID' and temp.text == str(TestCase_Id[j].text).split('-')[1]:
ATC_Name.next()
ATC_Name.next()
ATC_Name.next().text = 'Pass' <--This is what I want to do
ATC_Name.close()
break
except:
print sys.exc_info()
I believe my approach is wrong. Kindly guide me with right pointers.
Thanks.

You'd better switch to lxml so that you can use the "unlimited" power of xpath.
The idea is to use the following xpath expression:
//Procedure[ProcedureID/text()="%d"]/Verdict
where %d placeholder is substituted with the appropriate procedure id via string formatting operation.
The xpath expression finds the appropriate Verdict tag which you can set text on:
from lxml import etree
data = """<Procedure>
<PreCondition>PRECONDITION: - ECU in extended diagnostic session (zz = 0x03) </PreCondition>
<PostCondition/>
<ProcedureID>428495</ProcedureID>
<SequenceNumber>2</SequenceNumber>
<CID>-1</CID>
<Verdict Writable="true">NotExecuted</Verdict>
</Procedure>"""
ID = 428495
tree = etree.fromstring(data)
verdict = tree.xpath('//Procedure[ProcedureID/text()="%d"]/Verdict' % ID)[0]
verdict.text = 'test'
print etree.tostring(tree)
prints:
<Procedure>
<PreCondition>PRECONDITION: - ECU in extended diagnostic session (zz = 0x03) </PreCondition>
<PostCondition/>
<ProcedureID>428495</ProcedureID>
<SequenceNumber>2</SequenceNumber>
<CID>-1</CID>
<Verdict Writable="true">test</Verdict>
</Procedure>

Here is a solution using ElementTree. See Modifying an XML File
import xml.etree.ElementTree as et
tree = et.parse('prison.xml')
root = tree.getroot()
print root.find('Verdict').text #before update
root.find('Verdict').text = 'Executed'
tree.write('prison.xml')

try this
import xml.etree.ElementTree as et
root=et.parse(xmldata).getroot()
s=root.find('Verdict')
s.text='Your string'

xmlns namespace breaking lxml

I am trying to open an xml file, and get values from certain tags. I have done this a lot but this particular xml is giving me some issues. Here is a section of the xml file:
<?xml version='1.0' encoding='UTF-8'?>
<package xmlns="http://apple.com/itunes/importer" version="film4.7">
<provider>filmgroup</provider>
<language>en-GB</language>
<actor name="John Smith" display="Doe John"</actor>
</package>
And here is a sample of my python code:
metadata = '/Users/mylaptop/Desktop/Python/metadata.xml'
from lxml import etree
parser = etree.XMLParser(remove_blank_text=True)
open(metadata)
tree = etree.parse(metadata, parser)
root = tree.getroot()
for element in root.iter(tag='provider'):
providerValue = tree.find('//provider')
providerValue = providerValue.text
print providerValue
tree.write('/Users/mylaptop/Desktop/Python/metadataDone.xml', pretty_print = True, xml_declaration = True, encoding = 'UTF-8')
When I run this it can't find the provider tag or its value. If I remove xmlns="http://apple.com/itunes/importer" then all work as expected.
My question is how can I remove this namespace, as i'm not at all interested in this, so I can get the tag values I need using lxml?

The provider tag is in the http://apple.com/itunes/importer namespace, so you either need to use the fully qualified name
{http://apple.com/itunes/importer}provider
or use one of the lxml methods that has the namespaces parameter, such as root.xpath. Then you can specify it with a namespace prefix (e.g. ns:provider):
from lxml import etree
parser = etree.XMLParser(remove_blank_text=True)
tree = etree.parse(metadata, parser)
root = tree.getroot()
namespaces = {'ns':'http://apple.com/itunes/importer'}
items = iter(root.xpath('//ns:provider/text()|//ns:actor/#name',
namespaces=namespaces))
for provider, actor in zip(*[items]*2):
print(provider, actor)
yields
('filmgroup', 'John Smith')
Note that the XPath used above assumes that <provider> and <actor> elements always appear in alternation. If that is not true, then there are of course ways to handle it, but the code becomes a bit more verbose:
for package in root.xpath('//ns:package', namespaces=namespaces):
for provider in package.xpath('ns:provider', namespaces=namespaces):
providerValue = provider.text
print providerValue
for actor in package.xpath('ns:actor', namespaces=namespaces):
print actor.attrib['name']

My suggestion is to not ignore the namespace but, instead, to take it into account. I wrote some related functions (copied with slight modification) for my work on the django-quickbooks library. With these functions, you should be able to do this:
providers = getels(root, 'provider', ns='http://apple.com/itunes/importer')
Here are those functions:
def get_tag_with_ns(tag_name, ns):
return '{%s}%s' % (ns, tag_name)
def getel(elt, tag_name, ns=None):
""" Gets the first tag that matches the specified tag_name taking into
account the QB namespace.
:param ns: The namespace to use if not using the default one for
django-quickbooks.
:type ns: string
"""
res = elt.find(get_tag_with_ns(tag_name, ns=ns))
if res is None:
raise TagNotFound('Could not find tag by name "%s"' % tag_name)
return res
def getels(elt, *path, **kwargs):
""" Gets the first set of elements found at the specified path.
Example:
>>> xml = (
"<root>" +
"<item>" +
"<id>1</id>" +
"</item>" +
"<item>" +
"<id>2</id>"* +
"</item>" +
"</root>")
>>> el = etree.fromstring(xml)
>>> getels(el, 'root', 'item', ns='correct/namespace')
[<Element item>, <Element item>]
"""
ns = kwargs['ns']
i=-1
for i in range(len(path)-1):
elt = getel(elt, path[i], ns=ns)
tag_name = path[i+1]
return elt.findall(get_tag_with_ns(tag_name, ns=ns))

LXML Xpath does not seem to return full path

OK I'll be the first to admit its is, just not the path I want and I don't know how to get it.
I'm using Python 3.3 in Eclipse with Pydev plugin in both Windows 7 at work and ubuntu 13.04 at home. I'm new to python and have limited programming experience.
I'm trying to write a script to take in an XML Lloyds market insurance message, find all the tags and dump them in a .csv where we can easily update them and then reimport them to create an updated xml.
I have managed to do all of that except when I get all the tags it only gives the tag name and not the tags above it.
<TechAccount Sender="broker" Receiver="insurer">
<UUId>2EF40080-F618-4FF7-833C-A34EA6A57B73</UUId>
<BrokerReference>HOY123/456</BrokerReference>
<ServiceProviderReference>2012080921401A1</ServiceProviderReference>
<CreationDate>2012-08-10</CreationDate>
<AccountTransactionType>premium</AccountTransactionType>
<GroupReference>2012080921401A1</GroupReference>
<ItemsInGroupTotal>
<Count>1</Count>
</ItemsInGroupTotal>
<ServiceProviderGroupReference>8-2012-08-10</ServiceProviderGroupReference>
<ServiceProviderGroupItemsTotal>
<Count>13</Count>
</ServiceProviderGroupItemsTotal>
That is a fragment of the XML. What I want is to find all the tags and their path. For example for I want to show it as ItemsInGroupTotal/Count but can only get it as Count.
Here is my code:
xml = etree.parse(fullpath)
print( xml.xpath('.//*'))
all_xpath = xml.xpath('.//*')
every_tag = []
for i in all_xpath:
single_tag = '%s,%s' % (i.tag, i.text)
every_tag.append(single_tag)
print(every_tag)
This gives:
'{http://www.ACORD.org/standards/Jv-Ins-Reinsurance/1}ServiceProviderGroupReference,8-2012-08-10', '{http://www.ACORD.org/standards/Jv-Ins-Reinsurance/1}ServiceProviderGroupItemsTotal,\n', '{http://www.ACORD.org/standards/Jv-Ins-Reinsurance/1}Count,13',
As you can see Count is shown as {namespace}Count, 13 and not {namespace}ItemsInGroupTotal/Count, 13
Can anyone point me towards what I need?
Thanks (hope my first post is OK)
Adam
EDIT:
This is my code now:
with open(fullpath, 'rb') as xmlFilepath:
xmlfile = xmlFilepath.read()
fulltext = '%s' % xmlfile
text = fulltext[2:]
print(text)
xml = etree.fromstring(fulltext)
tree = etree.ElementTree(xml)
every_tag = ['%s, %s' % (tree.getpath(e), e.text) for e in xml.iter()]
print(every_tag)
But this returns an error:
ValueError: Unicode strings with encoding declaration are not supported. Please use bytes input or XML fragments without declaration.
I remove the first two chars as thy are b' and it complained it didn't start with a tag
Update:
I have been playing around with this and if I remove the xis: xxx tags and the namespace stuff at the top it works as expected. I need to keep the xis tags and be able to identify them as xis tags so can't just delete them.
Any help on how I can achieve this?

ElementTree objects have a method getpath(element), which returns a
structural, absolute XPath expression to find that element
Calling getpath on each element in a iter() loop should work for you:
from pprint import pprint
from lxml import etree
text = """
<TechAccount Sender="broker" Receiver="insurer">
<UUId>2EF40080-F618-4FF7-833C-A34EA6A57B73</UUId>
<BrokerReference>HOY123/456</BrokerReference>
<ServiceProviderReference>2012080921401A1</ServiceProviderReference>
<CreationDate>2012-08-10</CreationDate>
<AccountTransactionType>premium</AccountTransactionType>
<GroupReference>2012080921401A1</GroupReference>
<ItemsInGroupTotal>
<Count>1</Count>
</ItemsInGroupTotal>
<ServiceProviderGroupReference>8-2012-08-10</ServiceProviderGroupReference>
<ServiceProviderGroupItemsTotal>
<Count>13</Count>
</ServiceProviderGroupItemsTotal>
</TechAccount>
"""
xml = etree.fromstring(text)
tree = etree.ElementTree(xml)
every_tag = ['%s, %s' % (tree.getpath(e), e.text) for e in xml.iter()]
pprint(every_tag)
prints:
['/TechAccount, \n',
'/TechAccount/UUId, 2EF40080-F618-4FF7-833C-A34EA6A57B73',
'/TechAccount/BrokerReference, HOY123/456',
'/TechAccount/ServiceProviderReference, 2012080921401A1',
'/TechAccount/CreationDate, 2012-08-10',
'/TechAccount/AccountTransactionType, premium',
'/TechAccount/GroupReference, 2012080921401A1',
'/TechAccount/ItemsInGroupTotal, \n',
'/TechAccount/ItemsInGroupTotal/Count, 1',
'/TechAccount/ServiceProviderGroupReference, 8-2012-08-10',
'/TechAccount/ServiceProviderGroupItemsTotal, \n',
'/TechAccount/ServiceProviderGroupItemsTotal/Count, 13']
UPD:
If your xml data is in the file test.xml, the code would look like:
from pprint import pprint
from lxml import etree
xml = etree.parse('test.xml').getroot()
tree = etree.ElementTree(xml)
every_tag = ['%s, %s' % (tree.getpath(e), e.text) for e in xml.iter()]
pprint(every_tag)
Hope that helps.

getpath() does indeed return an xpath that's not suited for human consumption. From this xpath, you can build up a more useful one though. Such as with this quick-and-dirty approach:
def human_xpath(element):
full_xpath = element.getroottree().getpath(element)
xpath = ''
human_xpath = ''
for i, node in enumerate(full_xpath.split('/')[1:]):
xpath += '/' + node
element = element.xpath(xpath)[0]
namespace, tag = element.tag[1:].split('}', 1)
if element.getparent() is not None:
nsmap = {'ns': namespace}
same_name = element.getparent().xpath('./ns:' + tag,
namespaces=nsmap)
if len(same_name) > 1:
tag += '[{}]'.format(same_name.index(element) + 1)
human_xpath += '/' + tag
return human_xpath