I use xmltodict to convert Python dictionaries to XML. I would like to include the namespaces. What I've come up with is this:
xml_dict = {
"http://namespace.org:workflow":
{"action-list": "..."}
}
namespaces = {"http://namespace.org": "ws"}
xml = xmltodict.unparse(xml_dict, namespaces=namespaces,
pretty=True, short_empty_elements=True, full_document=False)
That gives me the result:
<ws:workflow>
<action-list>...</action-list>
</ws:workflow>
How can I include the namespaces in the result? I would like something like this:
<ws:workflow xmlns:ws="http://namespace.org">
Is there a simple solution using xmltodict?
Note: I've checked all keyword arguments in the source code of unparse and _emit functions. I guess the preprocessor argument will be the key to a better solution. But it's not documented.
Not nice, and not a general solution to add all the namespaces, but works.
import xmltodict
xml_dict = {
"http://namespace.org:workflow":
{
"#REPLACE_THIS": "http://namespace.org",
"action-list": "..."
}
}
namespaces = {"http://namespace.org": "ws"}
xml = xmltodict.unparse(xml_dict, namespaces=namespaces, pretty=True, short_empty_elements=True, full_document=False)
xml = xml.replace("REPLACE_THIS", "xmlns:ws", 1)
print(xml)
That gives me exactly what I want:
<ws:workflow xmlns:ws="http://namespace.org">
<action-list>...</action-list>
</ws:workflow>
The third argument in the replace function makes it sure that not other occurrences of "REPLACE_THIS" will be replaced, just the first one.
Related
Good Morning,
I need to understand how to insert a variable into this variable (CHANGEME).
payload = "{\n\t"client": {\n\t\t"clientId": "name"\n\t},\n\t"contentFieldOption": {\n\t\t"returnLinkedContents": false,\n\t\t"returnLinkedCategories": false,\n\t\t"returnEmbedCodes": false,\n\t\t"returnThumbnailUrl": false,\n\t\t"returnItags": false,\n\t\t"returnAclInfo": false,\n\t\t"returnImetadata": false,\n\t\t"ignoreITagCombining": false,\n\t\t"returnTotalResults": true\n\t},\n\t"criteria": {\n\t\t"linkedCategoryOp": {\n\t\t\t"linkedCategoryIds": [\n\t\t\t\t" CHANGEME ",\n\t\t\t\t"!_TRASH"\n\t\t\t],\n\t\t\t"cascade": true\n\t\t}\n\t},\n\t"numberOfresults": 50,\n\t"offset": 0,\n\t"orderBy": "creationDate_A"\n}"
It is part of the body to be inserted inside API POST request.
I have tried various alternatives, but to no avail it led me to solve my problem
Don't try to hack this string with regexes; you'll end up with invalid data in no time. Use json.loads() to convert it into a dictionary, find the key CHANGEME, and do whatever you need to do (which you do not really explain).
>>> paydict = json.loads(payload)
>>> print(json.dumps(paydict, indent=4)
{
"criteria": {
"linkedCategoryOp": {
"linkedCategoryIds": [
" CHANGEME ",
"!_TRASH"
...
API objects usually have a consistent structure, so your variable is probably always in the list paydict["criteria"]["linkedCategoryOp"]["linkedCategoryIds"]. Find the index of " CHANGEME " in this list, and take it from there.
You can use re - Python's regular expressions module :
import re
payload = '{\n\t"client": {\n\t\t"clientId": "name"\n\t},\n\t"contentFieldOption": {\n\t\t"returnLinkedContents": false,\n\t\t"returnLinkedCategories": false,\n\t\t"returnEmbedCodes": false,\n\t\t"returnThumbnailUrl": false,\n\t\t"returnItags": false,\n\t\t"returnAclInfo": false,\n\t\t"returnImetadata": false,\n\t\t"ignoreITagCombining": false,\n\t\t"returnTotalResults": true\n\t},\n\t"criteria": {\n\t\t"linkedCategoryOp": {\n\t\t\t"linkedCategoryIds": [\n\t\t\t\t" CHANGEME ",\n\t\t\t\t"!_TRASH"\n\t\t\t],\n\t\t\t"cascade": true\n\t\t}\n\t},\n\t"numberOfresults": 50,\n\t"offset": 0,\n\t"orderBy": "creationDate_A"\n}'
payload = re.sub("\n|\t","",payload).strip() # do some cleanup
payload = re.sub("\s+CHANGEME\s+","NEW VALUE",payload) # Replace the value
print(payload) # CHANGEME is replaced with NEW VALUE
You could use a simple string replace to swap "CHANGEME" with something else.
new_str = 'IMCHANGED'
payload.replace('CHANGEME', new_str)
This solves your stated problem, unless there are extra constraints about what the payload looks like (right now you're assuming it's a string, or how many times the word CHANGEME occurs). Please clarify if that is the case.
I'm stumped with how to do the ElementTree namespace dictionary and subsequent find() and findall() calls using the documented sytnax:
A better way to search the namespaced XML example is to create a
dictionary with your own prefixes and use those in the search
functions:
ns = {'real_person': 'http://people.example.com',
'role': 'http://characters.example.com'}
for actor in root.findall('real_person:actor', ns):
name = actor.find('real_person:name', ns)
print(name.text)
for char in actor.findall('role:character', ns):
print(' |-->', char.text)
The issue i'm having is if i try to use the syntax noted in that doc, by passing the "ns" dictionary as a 2nd argument in find() or findall(), i get an empty list. If I type out the full namespace without passing the 2nd argument, it returns all of the expected elements.
I've defined my namespace dictionary as such:
ns = {'ws':'{urn:com.workday/workersync}'}
And here is the ElementTree and root setup:
xmlparser = ET.parse(xmlfile)
xmlroot = xmlparser.getroot()
Here is what i get when i try to use the dictionary shortcut syntax noted in the docs:
>>> xmlroot.findall('ws:Worker', ns)
[]
Just an empty list... Here is what i get if type out the namespace in the call:
xmlroot.findall('{urn:com.workday/workersync}Worker')
[<Element '{urn:com.workday/workersync}Worker' at 0x03220A78>, <Element'{urn:com.workday/workersync}Worker' at 0x0322D8C0>]
That returns the expected 2 elements in my sample file.
Here is what the top of my sample file looks like for reference:
<?xml version="1.0" encoding="UTF-8"?>
<ws:Worker_Sync xmlns:ws="urn:com.workday/workersync" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<ws:Header>
<ws:Version>34.0</ws:Version>
<ws:Prior_Entry_Time>2020-07-04T21:40:25.822-07:00</ws:Prior_Entry_Time>
<ws:Current_Entry_Time>2020-07-04T22:03:47.458-07:00</ws:Current_Entry_Time>
<ws:Prior_Effective_Time>2020-07-04T00:00:00.000-07:00</ws:Prior_Effective_Time>
<ws:Current_Effective_Time>2020-07-05T00:00:00.000-07:00</ws:Current_Effective_Time>
<ws:Full_File>true</ws:Full_File>
<ws:Document_Retention_Policy>30</ws:Document_Retention_Policy>
<ws:Worker_Count>2</ws:Worker_Count>
</ws:Header>
<ws:Worker>
*<snipped rest of XML data>*
The snipped XML data contains 2 <ws:Worker> elements with many subchildren under them.
I've been messing with this for longer than i'd care to admit. I feel like I'm missing something incredibly obvious, as to my eyes, my code looks like every example i've found online and the example code on the docs.
Please help!
Remove the curly brackets from the URI string. The namespace dictionary should look like this:
ns = {'ws': 'urn:com.workday/workersync'}
Another option is to use a wildcard for the namespace. This is supported for find() and findall() since Python 3.8:
print(xmlroot.findall('{*}Worker'))
Output:
[<Element '{urn:com.workday/workersync}Worker' at 0x033E6AC8>]
I am writing a script that processes a rdf:skos file with python3 and lxml:
I learnt that I need to pass to the findall procedure the namespaces that the XML mentions. (Ok, strange, since the XML files lists these in the header, so this seems like an unnecessary step but anyway).
When calling
for concept in root.findall('.//skos:Concept', namespaces=root.nsmap):
that works, because a root.nsmap is constructed by lxml.
But then later in my code I also need to perform a test on xml:lang
for pl in concept.findall(".//skos:prefLabel[#xml:lang='en']", namespaces=root.nsmap):
and here python tells me
SyntaxError: prefix 'xml' not found in prefix map
Ok, true, in my skos file there is no extra declaration for the xml namespace. So I try to add it to the root.nsmap dict
root.nsmap['xml'] = "http://www.w3.org/XML/1998/namespace"
but that too doesn't work
nsmap = {'rdf': 'http://www.w3.org/1999/02/22-rdf-syntax-ns#', 'uneskos': 'http://purl.org/umu/uneskos#', 'iso-thes': 'http://purl.org/iso25964/skos-thes#', 'dcterms': 'http://purl.org/dc/terms/', 'skos': 'http://www.w3.org/2004/02/skos/core#', 'rdfs': 'http://www.w3.org/2000/01/rdf-schema#'}
Seems I am not allowed to modify the root.nsmap?
Anyone an idea how this is done? I have processed tons of XML in the past with Perl XML::Twig which is very very comfortable and I assmue, the Python community has (at least) similarly comfortable ways to do that ... but how?
Any hint appreciated.
Modifying root.nsmap has no effect. But you can create another dictionary and modify that one. Example:
from lxml import etree
doc = """
<root xmlns:skos="http://www.w3.org/2004/02/skos/core#">
<skos:prefLabel xml:lang='en'>FOO</skos:prefLabel>
<skos:prefLabel xml:lang='de'>BAR</skos:prefLabel>
</root>"""
root = etree.fromstring(doc)
nsmap = root.nsmap
nsmap["xml"] = "http://www.w3.org/XML/1998/namespace"
en = root.find(".//skos:prefLabel[#xml:lang='en']", namespaces=nsmap)
print(en.text)
Output:
FOO
I have the following XML which I want to parse using Python's ElementTree:
<rdf:RDF xml:base="http://dbpedia.org/ontology/"
xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
xmlns:owl="http://www.w3.org/2002/07/owl#"
xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
xmlns="http://dbpedia.org/ontology/">
<owl:Class rdf:about="http://dbpedia.org/ontology/BasketballLeague">
<rdfs:label xml:lang="en">basketball league</rdfs:label>
<rdfs:comment xml:lang="en">
a group of sports teams that compete against each other
in Basketball
</rdfs:comment>
</owl:Class>
</rdf:RDF>
I want to find all owl:Class tags and then extract the value of all rdfs:label instances inside them. I am using the following code:
tree = ET.parse("filename")
root = tree.getroot()
root.findall('owl:Class')
Because of the namespace, I am getting the following error.
SyntaxError: prefix 'owl' not found in prefix map
I tried reading the document at http://effbot.org/zone/element-namespaces.htm but I am still not able to get this working since the above XML has multiple nested namespaces.
Kindly let me know how to change the code to find all the owl:Class tags.
You need to give the .find(), findall() and iterfind() methods an explicit namespace dictionary:
namespaces = {'owl': 'http://www.w3.org/2002/07/owl#'} # add more as needed
root.findall('owl:Class', namespaces)
Prefixes are only looked up in the namespaces parameter you pass in. This means you can use any namespace prefix you like; the API splits off the owl: part, looks up the corresponding namespace URL in the namespaces dictionary, then changes the search to look for the XPath expression {http://www.w3.org/2002/07/owl}Class instead. You can use the same syntax yourself too of course:
root.findall('{http://www.w3.org/2002/07/owl#}Class')
Also see the Parsing XML with Namespaces section of the ElementTree documentation.
If you can switch to the lxml library things are better; that library supports the same ElementTree API, but collects namespaces for you in .nsmap attribute on elements and generally has superior namespaces support.
Here's how to do this with lxml without having to hard-code the namespaces or scan the text for them (as Martijn Pieters mentions):
from lxml import etree
tree = etree.parse("filename")
root = tree.getroot()
root.findall('owl:Class', root.nsmap)
UPDATE:
5 years later I'm still running into variations of this issue. lxml helps as I showed above, but not in every case. The commenters may have a valid point regarding this technique when it comes merging documents, but I think most people are having difficulty simply searching documents.
Here's another case and how I handled it:
<?xml version="1.0" ?><Tag1 xmlns="http://www.mynamespace.com/prefix">
<Tag2>content</Tag2></Tag1>
xmlns without a prefix means that unprefixed tags get this default namespace. This means when you search for Tag2, you need to include the namespace to find it. However, lxml creates an nsmap entry with None as the key, and I couldn't find a way to search for it. So, I created a new namespace dictionary like this
namespaces = {}
# response uses a default namespace, and tags don't mention it
# create a new ns map using an identifier of our choice
for k,v in root.nsmap.iteritems():
if not k:
namespaces['myprefix'] = v
e = root.find('myprefix:Tag2', namespaces)
Note: This is an answer useful for Python's ElementTree standard library without using hardcoded namespaces.
To extract namespace's prefixes and URI from XML data you can use ElementTree.iterparse function, parsing only namespace start events (start-ns):
>>> from io import StringIO
>>> from xml.etree import ElementTree
>>> my_schema = u'''<rdf:RDF xml:base="http://dbpedia.org/ontology/"
... xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
... xmlns:owl="http://www.w3.org/2002/07/owl#"
... xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
... xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
... xmlns="http://dbpedia.org/ontology/">
...
... <owl:Class rdf:about="http://dbpedia.org/ontology/BasketballLeague">
... <rdfs:label xml:lang="en">basketball league</rdfs:label>
... <rdfs:comment xml:lang="en">
... a group of sports teams that compete against each other
... in Basketball
... </rdfs:comment>
... </owl:Class>
...
... </rdf:RDF>'''
>>> my_namespaces = dict([
... node for _, node in ElementTree.iterparse(
... StringIO(my_schema), events=['start-ns']
... )
... ])
>>> from pprint import pprint
>>> pprint(my_namespaces)
{'': 'http://dbpedia.org/ontology/',
'owl': 'http://www.w3.org/2002/07/owl#',
'rdf': 'http://www.w3.org/1999/02/22-rdf-syntax-ns#',
'rdfs': 'http://www.w3.org/2000/01/rdf-schema#',
'xsd': 'http://www.w3.org/2001/XMLSchema#'}
Then the dictionary can be passed as argument to the search functions:
root.findall('owl:Class', my_namespaces)
I've been using similar code to this and have found it's always worth reading the documentation... as usual!
findall() will only find elements which are direct children of the current tag. So, not really ALL.
It might be worth your while trying to get your code working with the following, especially if you're dealing with big and complex xml files so that that sub-sub-elements (etc.) are also included.
If you know yourself where elements are in your xml, then I suppose it'll be fine! Just thought this was worth remembering.
root.iter()
ref: https://docs.python.org/3/library/xml.etree.elementtree.html#finding-interesting-elements
"Element.findall() finds only elements with a tag which are direct children of the current element. Element.find() finds the first child with a particular tag, and Element.text accesses the element’s text content. Element.get() accesses the element’s attributes:"
To get the namespace in its namespace format, e.g. {myNameSpace}, you can do the following:
root = tree.getroot()
ns = re.match(r'{.*}', root.tag).group(0)
This way, you can use it later on in your code to find nodes, e.g using string interpolation (Python 3).
link = root.find(f"{ns}link")
This is basically Davide Brunato's answer however I found out that his answer had serious problems the default namespace being the empty string, at least on my python 3.6 installation. The function I distilled from his code and that worked for me is the following:
from io import StringIO
from xml.etree import ElementTree
def get_namespaces(xml_string):
namespaces = dict([
node for _, node in ElementTree.iterparse(
StringIO(xml_string), events=['start-ns']
)
])
namespaces["ns0"] = namespaces[""]
return namespaces
where ns0 is just a placeholder for the empty namespace and you can replace it by any random string you like.
If I then do:
my_namespaces = get_namespaces(my_schema)
root.findall('ns0:SomeTagWithDefaultNamespace', my_namespaces)
It also produces the correct answer for tags using the default namespace as well.
My solution is based on #Martijn Pieters' comment:
register_namespace only influences serialisation, not search.
So the trick here is to use different dictionaries for serialization and for searching.
namespaces = {
'': 'http://www.example.com/default-schema',
'spec': 'http://www.example.com/specialized-schema',
}
Now, register all namespaces for parsing and writing:
for name, value in namespaces.iteritems():
ET.register_namespace(name, value)
For searching (find(), findall(), iterfind()) we need a non-empty prefix. Pass these functions a modified dictionary (here I modify the original dictionary, but this must be made only after the namespaces are registered).
self.namespaces['default'] = self.namespaces['']
Now, the functions from the find() family can be used with the default prefix:
print root.find('default:myelem', namespaces)
but
tree.write(destination)
does not use any prefixes for elements in the default namespace.
I have the following XML which I want to parse using Python's ElementTree:
<rdf:RDF xml:base="http://dbpedia.org/ontology/"
xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
xmlns:owl="http://www.w3.org/2002/07/owl#"
xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
xmlns="http://dbpedia.org/ontology/">
<owl:Class rdf:about="http://dbpedia.org/ontology/BasketballLeague">
<rdfs:label xml:lang="en">basketball league</rdfs:label>
<rdfs:comment xml:lang="en">
a group of sports teams that compete against each other
in Basketball
</rdfs:comment>
</owl:Class>
</rdf:RDF>
I want to find all owl:Class tags and then extract the value of all rdfs:label instances inside them. I am using the following code:
tree = ET.parse("filename")
root = tree.getroot()
root.findall('owl:Class')
Because of the namespace, I am getting the following error.
SyntaxError: prefix 'owl' not found in prefix map
I tried reading the document at http://effbot.org/zone/element-namespaces.htm but I am still not able to get this working since the above XML has multiple nested namespaces.
Kindly let me know how to change the code to find all the owl:Class tags.
You need to give the .find(), findall() and iterfind() methods an explicit namespace dictionary:
namespaces = {'owl': 'http://www.w3.org/2002/07/owl#'} # add more as needed
root.findall('owl:Class', namespaces)
Prefixes are only looked up in the namespaces parameter you pass in. This means you can use any namespace prefix you like; the API splits off the owl: part, looks up the corresponding namespace URL in the namespaces dictionary, then changes the search to look for the XPath expression {http://www.w3.org/2002/07/owl}Class instead. You can use the same syntax yourself too of course:
root.findall('{http://www.w3.org/2002/07/owl#}Class')
Also see the Parsing XML with Namespaces section of the ElementTree documentation.
If you can switch to the lxml library things are better; that library supports the same ElementTree API, but collects namespaces for you in .nsmap attribute on elements and generally has superior namespaces support.
Here's how to do this with lxml without having to hard-code the namespaces or scan the text for them (as Martijn Pieters mentions):
from lxml import etree
tree = etree.parse("filename")
root = tree.getroot()
root.findall('owl:Class', root.nsmap)
UPDATE:
5 years later I'm still running into variations of this issue. lxml helps as I showed above, but not in every case. The commenters may have a valid point regarding this technique when it comes merging documents, but I think most people are having difficulty simply searching documents.
Here's another case and how I handled it:
<?xml version="1.0" ?><Tag1 xmlns="http://www.mynamespace.com/prefix">
<Tag2>content</Tag2></Tag1>
xmlns without a prefix means that unprefixed tags get this default namespace. This means when you search for Tag2, you need to include the namespace to find it. However, lxml creates an nsmap entry with None as the key, and I couldn't find a way to search for it. So, I created a new namespace dictionary like this
namespaces = {}
# response uses a default namespace, and tags don't mention it
# create a new ns map using an identifier of our choice
for k,v in root.nsmap.iteritems():
if not k:
namespaces['myprefix'] = v
e = root.find('myprefix:Tag2', namespaces)
Note: This is an answer useful for Python's ElementTree standard library without using hardcoded namespaces.
To extract namespace's prefixes and URI from XML data you can use ElementTree.iterparse function, parsing only namespace start events (start-ns):
>>> from io import StringIO
>>> from xml.etree import ElementTree
>>> my_schema = u'''<rdf:RDF xml:base="http://dbpedia.org/ontology/"
... xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
... xmlns:owl="http://www.w3.org/2002/07/owl#"
... xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
... xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
... xmlns="http://dbpedia.org/ontology/">
...
... <owl:Class rdf:about="http://dbpedia.org/ontology/BasketballLeague">
... <rdfs:label xml:lang="en">basketball league</rdfs:label>
... <rdfs:comment xml:lang="en">
... a group of sports teams that compete against each other
... in Basketball
... </rdfs:comment>
... </owl:Class>
...
... </rdf:RDF>'''
>>> my_namespaces = dict([
... node for _, node in ElementTree.iterparse(
... StringIO(my_schema), events=['start-ns']
... )
... ])
>>> from pprint import pprint
>>> pprint(my_namespaces)
{'': 'http://dbpedia.org/ontology/',
'owl': 'http://www.w3.org/2002/07/owl#',
'rdf': 'http://www.w3.org/1999/02/22-rdf-syntax-ns#',
'rdfs': 'http://www.w3.org/2000/01/rdf-schema#',
'xsd': 'http://www.w3.org/2001/XMLSchema#'}
Then the dictionary can be passed as argument to the search functions:
root.findall('owl:Class', my_namespaces)
I've been using similar code to this and have found it's always worth reading the documentation... as usual!
findall() will only find elements which are direct children of the current tag. So, not really ALL.
It might be worth your while trying to get your code working with the following, especially if you're dealing with big and complex xml files so that that sub-sub-elements (etc.) are also included.
If you know yourself where elements are in your xml, then I suppose it'll be fine! Just thought this was worth remembering.
root.iter()
ref: https://docs.python.org/3/library/xml.etree.elementtree.html#finding-interesting-elements
"Element.findall() finds only elements with a tag which are direct children of the current element. Element.find() finds the first child with a particular tag, and Element.text accesses the element’s text content. Element.get() accesses the element’s attributes:"
To get the namespace in its namespace format, e.g. {myNameSpace}, you can do the following:
root = tree.getroot()
ns = re.match(r'{.*}', root.tag).group(0)
This way, you can use it later on in your code to find nodes, e.g using string interpolation (Python 3).
link = root.find(f"{ns}link")
This is basically Davide Brunato's answer however I found out that his answer had serious problems the default namespace being the empty string, at least on my python 3.6 installation. The function I distilled from his code and that worked for me is the following:
from io import StringIO
from xml.etree import ElementTree
def get_namespaces(xml_string):
namespaces = dict([
node for _, node in ElementTree.iterparse(
StringIO(xml_string), events=['start-ns']
)
])
namespaces["ns0"] = namespaces[""]
return namespaces
where ns0 is just a placeholder for the empty namespace and you can replace it by any random string you like.
If I then do:
my_namespaces = get_namespaces(my_schema)
root.findall('ns0:SomeTagWithDefaultNamespace', my_namespaces)
It also produces the correct answer for tags using the default namespace as well.
My solution is based on #Martijn Pieters' comment:
register_namespace only influences serialisation, not search.
So the trick here is to use different dictionaries for serialization and for searching.
namespaces = {
'': 'http://www.example.com/default-schema',
'spec': 'http://www.example.com/specialized-schema',
}
Now, register all namespaces for parsing and writing:
for name, value in namespaces.iteritems():
ET.register_namespace(name, value)
For searching (find(), findall(), iterfind()) we need a non-empty prefix. Pass these functions a modified dictionary (here I modify the original dictionary, but this must be made only after the namespaces are registered).
self.namespaces['default'] = self.namespaces['']
Now, the functions from the find() family can be used with the default prefix:
print root.find('default:myelem', namespaces)
but
tree.write(destination)
does not use any prefixes for elements in the default namespace.