I am looking at a piece of XML that I want to add a node in.
<profile>
<dog>1</dog>
<halfdog>0</halfdog>
<cat>545</cat>
<lions>0</lions>
<bird>23</bird>
<dino>0</dino>
<pineapples>2</pineapples>
<people>0</people>
</profile>
With the above XML, I'm able to insert XML nodes into it. However, I'm not able to insert it at exact locations.
Is there a way to find if I am next to a certain node, whether it be before or after. Say if I wanted to add <snail>2</snail> between the <dino>0</dino> and <pineapples>2</pineapples> nodes.
Using ElementTree how can I find what node is next to me? I'm asking about ElementTree or any standard Python library. Unfortunately, lxml is out of the question for me.
I believe its not doable using ElementTree, but you can do it using the standard python minidom:
# create snail element
snail = dom.createElement('snail')
snail_text = dom.createTextNode('2')
snail.appendChild(snail_text)
# add it in the right place
profile = dom.getElementsByTagName('profile')[0]
pineapples = dom.getElementsByTagName('pineapples')[0]
profile.insertBefore(snail, pineapples)
output:
<?xml version="1.0" ?><profile>
<dog>1</dog>
<halfdog>0</halfdog>
<cat>545</cat>
<lions>0</lions>
<bird>23</bird>
<dino>0</dino>
<snail>2</snail><pineapples>2</pineapples>
<people>0</people>
</profile>
If you know the parent element and the element to insert before, you can use the following method with ElementTree:
index = parentElem.getchildren().index(elemToInsertBefore)
parent.insert(index, newElement)
Related
For my case, I have to find few elements in the XML file and update their values using the text attribute. For that, I have to search xml element A, B and C. My project is using xml.etree and python language. Currently I am using:
self.get_root.findall(H/A/T)
self.get_root.findall(H/B/T)
self.get_root.findall(H/C/T)
The sample XML file:
<H><A><T>text-i-have-to-update</H></A></T>
<H><B><T>text-i-have-to-update</H></B></T>
<H><C><T>text-i-have-to-update</H></C></T>
As we can notice, only the middle element in the path is different. Is there a way to optimize the code using something like self.get_root.findall(H|(A,B,C)|T)? Any guidance in the right direction will do! Thanks!
I went through the similar question: XPath to select multiple tags but it didn't work for my case
Update: maybe regular expression inside the findall()?
The html in your question is malformed; assuming it's properly formatted (like below), try this:
import xml.etree.ElementTree as ET
data = """<root>
<H><A><T>text-i-have-to-update</T></A></H>
<H><B><T>text-i-have-to-update</T></B></H>
<H><C><T>text-i-have-to-update</T></C></H>
</root>"""
doc = ET.fromstring(data)
for item in doc.findall('.//H//T'):
item.text = "modified text"
print(ET.tostring(doc).decode())
Output:
<root>
<H><A><T>modified text</T></A></H>
<H><B><T>modified text</T></B></H>
<H><C><T>modified text</T></C></H>
</root>
Given the xml
xmlstr = '''
<myxml>
<Description id="10">
<child info="myurl"/>
</Description>
</myxml>'
I'd like to get the id of Description only where child has an attribute of info.
import xml.etree.ElementTree as ET
root = ET.fromstring(xmlstr)
a = root.find(".//Description/[child/#info]")
print(a.attrib)
and changing the find to .//Description/[child[#info]]
both return an error of:
SyntaxError: invalid predicate
I know that etree only supports a subset of xpath, but this doesn't seem particularly weird - should this work? If so, what have I done wrong?!
Changing the find to .//Description/[child] does work, and returns
{'id': '10'}
as expected
You've definitely hit that XPath limited support limitation as, if we look at the source directly (looking at 3.7 source code), we could see that while parsing the Element Path expression, only these things in the filters are considered:
[#attribute] predicate
[#attribute='value']
[tag]
[.='value'] or [tag='value']
[index] or [last()] or [last()-index]
Which means that both of your rather simple expressions are not supported.
If you really want/need to stick with the built-in ElementTree library, one way to solve this would be with finding all Description tags via .findall() and filtering the one having a child element with info attribute.
You can also get those values as keys, which makes it a bit more structured approach to gather data:
import xml.etree.ElementTree as ET
root = ET.fromstring(xmlstr)
wht =root.find(".//Description")
wht.keys() #--> ['id']
wht.get('id') # --> '10'
I am trying to debug some inherited code. So there is a line of code
elt = doc.xpath('body/div/pre[#id="bb flags"]')[0]
I want to see what the entire document looks like at that point, rather than just a specific piece. So what xpath should I insert there?
i.e.
elt_entire_document = doc.xpath(new xpath)
logging.info("The full document here is " + elt_entire_document.text)
Is this even possible with xpath, or is it more complicated than that?
Simply like this, based on our comments :
logging.info("The full document here is " + text)
Your question title seems to be asking about selecting a whole document, yet your question body seems to be asking about displaying a selected node...
Selecting the whole document via XPath
Selecting the whole document might mean any of the following:
/ selects the root node of XML document.
/* selects the document element (aka the root element) of the XML document.(Its parent is the root node.)
string(/) selects the string-value of the XML document.
See also:
What is the difference between root node, root element and document element in XML?
How to pretty print XML from the command line?
Python pretty print subtree
How to use lxml and python to pretty print a subtree of an xml file?
how to get the full contents of a node using xpath & lxml?
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)
...
I am a first time XPath user and need to be able to get the text values of these different elements.. for instance time, title, etc.. I am using the libxml2 module in Python and so far have not had much luck getting just the values of the text I need. The code below here only returns the element tags.. i need the values.. any help would be GREATLY appreciated!
I'm using this code:
doc = libxml2.parseDoc(xmlOutput)
result = doc.xpathEval('//*')
With the following document:
<?xml version="1.0" encoding="UTF-8" ?>
<!DOCTYPE SCAN_LIST_OUTPUT SYSTEM "https://qualysapi.qualys.com/api/2.0/fo/sca/scan_list_output.dtd">
<SCAN_LIST_OUTPUT>
<RESPONSE>
<DATETIME>2012-01-22T01:21:53Z</DATETIME>
<SCAN_LIST>
<SCAN>
<REF>scan/2343423</REF>
<TYPE>Scheduled</TYPE>
<TITLE><![CDATA[customer 1 5/20/2012]]></TITLE>
<USER_LOGIN>user1</USER_LOGIN>
<LAUNCH_DATETIME>2012-02-21T04:11:05Z</LAUNCH_DATETIME>
<STATUS>
<STATE>Finished</STATE>
</STATUS>
<TARGET><![CDATA[13.3.3.2, 13.8.8.10, 13.10.12.60, 13.10.12.11...]]></TARGET>
</SCAN>
</SCAN_LIST>
</RESPONSE>
</SCAN_LIST_OUTPUT>
You can call getContent() on each returned xmlNode object to retrieve the associated text. Note that this is recursive -- to non-recursively access text content in libxml2, you'll want to retrieve the associated text node under the element, and call .getContent() on that.
That said, this would be easier if you used lxml.etree (a higher-level Python API, still backing into the C libxml2 library) instead of the Python libxml2; in that case, it's simply element.text to access the associated content as a string.
Have a look at Mark Pilgrim's Dive Into Python 3, Chapter 12. XML
The chapter starts with short course to XML (general talk but with the Atom Syndication Feed example), then it continues with the standard xml.etree.ElementTree and continues with third party lxml that implements more with the same interface (full XPATH 1.0, based on libxml2).