python to pars xml to get value - python

I have a xml response from one of my system where i am trying to get the value using python code. Need experts view on highlighting my mistake.
<?xml version="1.0" encoding="UTF-8" standalone="yes"?><ns3:loginResponse xmlns:ns2="http://ws.core.product.xxxxx.com/groupService/" xmlns:ns3="http://ws.core.product.xxxxx.com/loginService/" xmlns:ns4="http://ws.core.product.xxxxx.com/userService/"><ns3:return>YWeVDwuZwHdxxxxxxxxxxx_GqLtkNTE.</ns3:return></ns3:loginResponse>
I am using the below code of code and had no luck in getting the value - YWeVDwuZwHdxxxxxxxxxxx_GqLtkNTE . I haven't used xml parsing with namespace. response.text has the above xml response.
responsetree = ET.ElementTree(ET.fromstring(response.text))
responseroot = responsetree.getroot()
for a in root.iter('return'):
print(a.attrib)

YWeVDwuZwHdxxxxxxxxxxx_GqLtkNTE is not in the attrib. It is the element text
The attrib in this case is an empty dict
See https://www.cmi.ac.in/~madhavan/courses/prog2-2012/docs/diveintopython3/xml.html about parsing XML dics using namespace.

Reference from other answer helped to understand the concepts.
Once I understood the xml structure , Its plain simple. Just adding the output it might help someone in future for quick reference.
responsetree = ET.ElementTree(ET.fromstring(response.text))
responseroot = responsetree.getroot()
root[0].text
Keeping it simple for understanding. You might need to find the len(root) and/or iterate over the loop with condition to get apt value. You can also use findall , find along with to get the interested item.

Related

Parsing Google Earth KML file in Python (lxml, namespaces)

I am trying to parse a .kml file into Python using the xml module (after failing to make this work in BeautifulSoup, which I use for HTML).
As this is my first time doing this, I followed the official tutorial and all goes well until I try to construct an iterator to extract my data by root iteration:
from lxml import etree
tree=etree.parse('kmlfile')
Here is the example from the tutorial I am trying to emulate:
If you know you are only interested in a single tag, you can pass its name to getiterator() to have it filter for you:
for element in root.getiterator("child"):
print element.tag, '-', element.text
I would like to get all data under 'Placemark', so I tried
for i in tree.getiterterator("Placemark"):
print i, type(i)
which doesn't give me anything. What does work is:
for i in tree.getiterterator("{http://www.opengis.net/kml/2.2}Placemark"):
print i, type(i)
I don't understand how this comes about. The www.opengis.net is listed in the tag at the beginning of the document (kml xmlns="http://www.opengis.net/kml/2.2"...) , but I don't understand
how the part in {} relates to my specific example at all
why it is different from the tutorial
and what I am doing wrong
Any help is much appreciated!
Here is my solution.
So, the most important thing to do is read this as posted by Tomalak. It's a really good description of namespaces and easy to understand.
We are going to use XPath to navigate the XML document. Its notation is similar to file systems, where parents and descendants are separated by slashes /. The syntax is explained here, but note that some commands are different for the lxml implementation.
###Problem
Our goal is to extract the city name: the content of <name> which is under <Placemark>. Here's the relevant XML:
<Placemark> <name>CITY NAME</name>
The XPath equivalent to the non-functional code I posted above is:
tree=etree.parse('kml document')
result=tree.xpath('//Placemark/name/text()')
Where the text() part is needed to get the text contained in the location //Placemark/name.
Now this doesn't work, as Tomalak pointed out, cause the name of these two nodes are actually {http://www.opengis.net/kml/2.2}Placemark and {http://www.opengis.net/kml/2.2}name. The part in curly brackets is the default namespace. It does not show up in the actual document (which confused me) but it is defined at the beginning of the XML document like this:
xmlns="http://www.opengis.net/kml/2.2"
###Solution
We can supply namespaces to xpath by setting the namespaces argument:
xpath(X, namespaces={prefix: namespace})
This is easy enough for the namespaces that have actual prefixes, in this document for instance <gx:altitudeMode>relativeToSeaFloor</gx:altitudeMode> where the gx prefix is defined in the document as xmlns:gx="http://www.google.com/kml/ext/2.2".
However, Xpath does not understand what a default namespace is (cf docs). Therefore, we need to trick it, like Tomalak suggested above: We invent a prefix for the default and add it to our search terms. We can just call it kml for instance. This piece of code actually does the trick:
tree.xpath('//kml:Placemark/kml:name/text()', namespaces={"kml":"http://www.opengis.net/kml/2.2"})
The tutorial mentions that there is also an ETXPath method, that works just like Xpath except that one writes the namespaces out in curly brackets instead of defining them in a dictionary. Thus, the input would be of the style {http://www.opengis.net/kml/2.2}Placemark.

using python to parse xml data

I have a question with regards to XML and python. I want to comb through this xml file and look for certain tags, and then within those tags look for where there is data separated by a comma. split that and make a new line. I have the logic down, im just not too familiar with python to know whoch modules I should be researching. Any help as to where i should start researching would help.
172.28.18.142,10.0.0.2
thanks
I think when it comes to xml parsing in python there are a few options: lxml, xml, and BeautifulSoup. Most of my experience has dealt with the first two and I've found lxml to be extraordinarily faster than xml. Here's an lxml code snippet for parsing all elements of the root with a particular tag and storing the comma-separated text of each tag as a list. I think you'll want to add a lot of try and except blocks and tinker with the details, but this should get you started.
from lxml import etree
file_path = r'C:\Desktop\some_file.xml'
tree = etree.parse(file_path)
info_list = []
my_tag_path = tree.xpath('//topTag')
for elem in my_tag_path:
if elem.find('.//childTag') is not None:
info_list.append(elem.xpath('.//childTag')[0].text.split(','))

python etree with xpath and namespaces with prefix

I can't find info, how to parse my XML with namespace:
I have this xml:
<par:Request xmlns:par="http://somewhere.net/actual">
<par:actual>blabla</par:actual>
<par:documentType>string</par:documentType>
</par:Request>
And tried to parse it:
dom = ET.parse(u'C:\\filepath\\1.xml')
rootxml = dom.getroot()
for subtag in rootxml.xpath(u'//par:actual'):
#do something
print(subtag)
And got exception, because it doesn't know about namespace prefix.
Is there best way to solve that problem, counting that script will not know about file it going to parse and tag is going to search for?
Searching web and stackoverflow I found, that if I will add there:
namespace = {u'par': u"http://somewhere.net/actual"}
for subtag in rootxml.xpath(u'//par:actual', namespaces=namespace):
#do something
print(subtag)
That works. Perfect. But I don't know which XML I will parse, and searching tag (such as //par:actual) is also unknown to my script. So, I need to find way to extract namespace from XML somehow.
I found a lot of ways, how to extract namespace URI, such as:
print(rootxml.tag)
print(rootxml.xpath('namespace-uri(.)'))
print(rootxml.xpath('namespace-uri(/*)'))
But how should I extract prefix to create dictionary which ElementTree wants from me? I don't want to use regular expression monster over xml body to extract prefix, I believe there have to exist supported way for that, isn't it?
And maybe there have to exist some methods for me to extract by ETree namespace from XML as dictionary (as ETree wants!) without hands manipulation?
You cannot rely on the namespace declarations on the root element: there is no guarantee that the declarations will even be there, or that the document will have the same prefix for the same namespace throughout.
Assuming you are going to have some way of passing the tag you want to search (because you say it is not known by your script), you should also provide a way to pass a namespace mapping as well. Or use the James Clark notation, like {http://somewhere.net/actual}actual (the ETXPath has support for this syntax, whereas "normal" xpath does not, but you can also use other methods like .findall() if you don't need full xpath)
If you don't care for the prefix at all, you could also use the local-name() function in xpath, eg. //*[local-name()="actual"] (but you won't be "really" sure it's the right "actual")
Oh, I found it.
After we do that:
dom = ET.parse(u'C:\\filepath\\1.xml')
rootxml = dom.getroot()
Object rootxml contains dictionary nsmap, which contains all namespaces that I want.
So, simplest solution I've found:
dom = ET.parse(u'C:\\filepath\\1.xml')
rootxml = dom.getroot()
nss = rootxml.nsmap
for subtag in rootxml.xpath(u'//par:actual', namespaces=nss):
#do something
print(subtag)
That works.
UPD: that works if user understand what means 'par' in XML he works with. For example, comparing supposed namespace with existing namespace before any other operations.
Still, I like much variant with XPath that understands {...}actual, that was what I tried to achieve.
With Python 3.8.2 I found this question with the same issue.
This is the solution I found, put the namespace in the XPath query. (Between the {})
ApplicationArea = BOD_IN_tree.find('.//ApplicationArea', ns)
if(ApplicationArea is None):
ApplicationArea = BOD_IN_tree.find('.//{http://www.defaultNamespace.com/2}ApplicationArea', ns)
I search for the element without the namespace, then search again if it's not found. I have no control over the inbound documents, some have namespaces, some do not.
I hope this helps!

Parsing XML data within tags using lxml in python

My question is regarding how to get information stored in a tag which allows for no closing tag. Here's the relevant xml:
<?xml version="1.0" encoding="UTF-8"?>
<uws:job>
<uws:results>
<uws:result id="2014-03-03T15:42:31:1337" xlink:href="http://www.cosmosim.org/query/index/stream/table/2014-03-03T15%3A42%3A31%3A1337/format/csv" xlink:type="simple"/>
</uws:results>
</uws:job>
I'm looking to extract the xlink:href url here. As you can see the uws:result tag requires no closing tag. Additionally, having the 'uws:' makes it a bit tricky to handle them when working in python. Here's what I've tried so far:
from lxml import etree
root = etree.fromstring(xmlresponse.content)
url = root.find('{*}results').text
Where xmlresponse.content is the xml data to be parsed. What this returns is
'\n '
which indicates that it's only finding the newline character, since what I'm really after is contained within a tag inside the results tag. Any ideas would be greatly appreciated.
You found the right node; you extracted the data incorrectly. Instead of
url = root.find('{*}results').text
you really want
url = root.find('{*}results').get('attribname', 'value_to_return_if_not_present')
or
url = root.find('{*}results').attrib['attribname']
(which will throw an exception if not present).
Because of the namespace on the attribute itself, you will probably need to use the {ns}attrib syntax to look it up too.
You can dump out the attrib dictionary and just copy the attribute name out too.
text is actually the space between elements, and is not normally used but is supported both for spacing (like etreeindent) and some special cases.

XML parsing in Python using Python 2 or 3

I'm just trying to write a simple program to allow me to parse some of the following XML.
So far in following examples I am not getting the results I'm looking for.
I encounter many of these XML files and I generally want the info after a handful of tags.
What's the best way using elementtree to be able to do a search for <Id> and grab what ever info is in that tag. I was trying things like
for Reel in root.findall('Reel'):
... id = Reel.findtext('Id')
... print id
Is there a way just to look for every instance of <Id> and grab the urn: etc that comes after it? Some code that traverses everything and looks for <what I want> and so on.
This is a very truncated version of what I usually deal with.
This didn't get what I wanted at all. Is there an easy just to match <what I want> in any XML file and get the contents of that tag, or do i need to know the structure of the XML well enough to know its relation to Root/child etc?
<Reel>
<Id>urn:uuid:632437bc-73f9-49ca-b687-fdb3f98f430c</Id>
<AssetList>
<MainPicture>
<Id>urn:uuid:46afe8a3-50be-4986-b9c8-34f4ba69572f</Id>
<EditRate>24 1</EditRate>
<IntrinsicDuration>340</IntrinsicDuration>
<EntryPoint>0</EntryPoint>
<Duration>340</Duration>
<FrameRate>24 1</FrameRate>
<ScreenAspectRatio>2048 858</ScreenAspectRatio>
</MainPicture>
<MainSound>
<Id>urn:uuid:1fce0915-f8c7-48a7-b023-36e204a66ed1</Id>
<EditRate>24 1</EditRate>
<IntrinsicDuration>340</IntrinsicDuration>
<EntryPoint>0</EntryPoint>
<Duration>340</Duration>
</MainSound>
</AssetList>
</Reel>
#Mata that worked perfectly, but when I tried to use that for different values on another XML file I fell flat on my face. For instance, what about this section of a file.I couldn't post the whole thing unfortunately. What if I want to grab what comes after KeyId?
<?xml version="1.0" encoding="UTF-8" standalone="no" ?><DCinemaSecurityMessage xmlns="http://www.digicine.com/PROTO-ASDCP-KDM-20040311#" xmlns:dsig="http://www.w3.org/2000/09/xmldsig#" xmlns:enc="http://www.w3.org/2001/04/xmlenc#">
<!-- Generated by Wailua Version 0.3.20 -->
<AuthenticatedPublic Id="ID_AuthenticatedPublic">
<MessageId>urn:uuid:7bc63f4c-c617-4d00-9e51-0c8cd6a4f59e</MessageId>
<MessageType>http://www.digicine.com/PROTO-ASDCP-KDM-20040311#</MessageType>
<AnnotationText>SPIDERMAN-3_FTR_S_EN-XX_US-13_51_4K_PH_20070423_DELUXE ~ KDM for Quvis-10010.pem</AnnotationText>
<IssueDate>2007-04-29T04:13:43-00:00</IssueDate>
<Signer>
<dsig:X509IssuerName>dnQualifier=BzC0n/VV/uVrl2PL3uggPJ9va7Q=,CN=.deluxe-admin-c,OU=.mxf-j2c.ca.cinecert.com,O=.ca.cinecert.com</dsig:X509IssuerName>
<dsig:X509SerialNumber>10039</dsig:X509SerialNumber>
</Signer>
<RequiredExtensions>
<Recipient>
<X509IssuerSerial>
<dsig:X509IssuerName>dnQualifier=RUxyQle0qS7qPbcNRFBEgVjw0Og=,CN=SM.QuVIS.com.001,OU=QuVIS Digital Cinema,O=QuVIS.com</dsig:X509IssuerName>
<dsig:X509SerialNumber>363</dsig:X509SerialNumber>
</X509IssuerSerial>
<X509SubjectName>CN=SM MD LE FM.QuVIS_CinemaPlayer-3d_10010,OU=QuVIS,O=QuVIS.com,dnQualifier=3oBfjTfx1me0p1ms7XOX\+eqUUtE=</X509SubjectName>
</Recipient>
<CompositionPlaylistId>urn:uuid:336263da-e4f1-324e-8e0c-ebea00ff79f4</CompositionPlaylistId>
<ContentTitleText>SPIDERMAN-3_FTR_S_EN-XX_US-13_51_4K_PH_20070423_DELUXE</ContentTitleText>
<ContentKeysNotValidBefore>2007-04-30T05:00:00-00:00</ContentKeysNotValidBefore>
<ContentKeysNotValidAfter>2007-04-30T10:00:00-00:00</ContentKeysNotValidAfter>
<KeyIdList>
<KeyId>urn:uuid:9851b0f6-4790-0d4c-a69d-ea8abdedd03d</KeyId>
<KeyId>urn:uuid:8317e8f3-1597-494d-9ed8-08a751ff8615</KeyId>
<KeyId>urn:uuid:5d9b228d-7120-344c-aefc-840cdd32bbfc</KeyId>
<KeyId>urn:uuid:1e32ccb2-ab0b-9d43-b879-1c12840c178b</KeyId>
<KeyId>urn:uuid:44d04416-676a-2e4f-8995-165de8cab78d</KeyId>
<KeyId>urn:uuid:906da0c1-b0cb-4541-b8a9-86476583cdc4</KeyId>
<KeyId>urn:uuid:0fe2d73a-ebe3-9844-b3de-4517c63c4b90</KeyId>
<KeyId>urn:uuid:862fa79a-18c7-9245-a172-486541bef0c0</KeyId>
<KeyId>urn:uuid:aa2f1a88-7a55-894d-bc19-42afca589766</KeyId>
<KeyId>urn:uuid:59d6eeff-cd56-6245-9f13-951554466626</KeyId>
<KeyId>urn:uuid:14a13b1a-76ba-764c-97d0-9900f58af53e</KeyId>
<KeyId>urn:uuid:ccdbe0ae-1c3f-224c-b450-947f43bbd640</KeyId>
<KeyId>urn:uuid:dcd37f10-b042-8e44-bef0-89bda2174842</KeyId>
<KeyId>urn:uuid:9dd7103e-7e5a-a840-a15f-f7d7fe699203</KeyId>
</KeyIdList>
</RequiredExtensions>
<NonCriticalExtensions/>
</AuthenticatedPublic>
<AuthenticatedPrivate Id="ID_AuthenticatedPrivate"><enc:EncryptedKey xmlns:enc="http://www.w3.org/2001/04/xmlenc#">
<enc:EncryptionMethod Algorithm="http://www.w3.org/2001/04/xmlenc#rsa-oaep-mgf1p">
<ds:DigestMethod xmlns:ds="http://www.w3.org/2000/09/xmldsig#" Algorithm="http://www.w3.org/2000/09/xmldsig#sha1"/>
</enc:EncryptionMethod>
The expression Reel.findtext('Id') only matches direct children of Reel. If you want to find all Id tags in your xml document, you can just use:
ids = [id.text for id in Reel.findall(".//Id")]
This would give you a list of all text nodes of all Id tags which are children of Reel.
edit:
Your updated example uses namespaces, in this case KeyId is in the default namespace (http://www.digicine.com/PROTO-ASDCP-KDM-20040311#), so to search for it you need to include it in your search:
from xml.etree import ElementTree
doc = ElementTree.parse('test.xml')
nsmap = {'ns': 'http://www.digicine.com/PROTO-ASDCP-KDM-20040311#'}
ids = [id.text for id in doc.findall(".//ns:KeyId", namespaces=nsmap)]
print(ids)
...
The xpath subset ElementTree supports is rather limited. If you want a more complete support, you should use lxml instead, it's xpath support is way more complete.
For example, using xpath to search for all KeyId tags (ignoring namespaces) and returning their text content directly:
from lxml import etree
doc = etree.parse('test.xml')
ids = doc.xpath(".//*[local-name()='KeyId']/text()")
print(ids)
...
It sounds like XPath might be right up your alley - it will let you query your XML document for exactly what you're looking for, as long as you know the structure.
Here's what I needed to do. This works for finding whatever I need.
for node in tree.getiterator():
... if 'KeyId' in node.tag:
... mylist = node.tag
... print(mylist)
...

Categories