I am trying to include a reference to a DTD in my XML doc using minidom.
I am creating the document like:
doc = Document()
foo = doc.createElement('foo')
doc.appendChild(foo)
doc.toxml()
This gives me:
<?xml version="1.0" ?>
<foo/>
I need to get something like:
<?xml version="1.0" ?>
<!DOCTYPE something SYSTEM "http://www.path.to.my.dtd.com/my.dtd">
<foo/>
The documentation is out of date. Use the source, Luke. I do it something like this.
from xml.dom.minidom import DOMImplementation
imp = DOMImplementation()
doctype = imp.createDocumentType(
qualifiedName='foo',
publicId='',
systemId='http://www.path.to.my.dtd.com/my.dtd',
)
doc = imp.createDocument(None, 'foo', doctype)
doc.toxml()
This prints the following.
<?xml version="1.0" ?><!DOCTYPE foo SYSTEM \'http://www.path.to.my.dtd.com/my.dtd\'><foo/>
Note how the root element is created automatically by createDocument(). Also, your 'something' has been changed to 'foo': the DTD needs to contain the root element name itself.
According to the Python docs, there is no implementation of the DocumentType interface in the minidom.
Related
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 am writing program to work on xml file and change it. But when I try to get to any part of it I get some extra part.
My xml file:
<?xml version="1.0" encoding="UTF-8"?>
<Package xmlns="http://soap.sforce.com/2006/04/metadata">
<types>
<members>sbaa__ApprovalChain__c.ExternalID__c</members>
<members>sbaa__ApprovalCondition__c.ExternalID__c</members>
<members>sbaa__ApprovalRule__c.ExternalID__c</members>
<name>CustomField</name>
</types>
<version>40.0</version>
</Package>
And I have my code:
from lxml import etree
import sys
tree = etree.parse('package.xml')
root = tree.getroot()
print( root[0][0].tag )
As output I expect to see members but I get something like this:
{http://soap.sforce.com/2006/04/metadata}members
Why do I see that url and how to stop it from showing up?
You have defined a default namespace (Wikipedia, lxml tutorial). When defined, it is a part of every child tag.
If you want to print the tag without the namespace, it's easy
tag = root[0][0].tag
print(tag[tag.find('}')+1:])
If you want to remove the namespace from XML, see this question.
So I have the following XML document It is much longer:
<?xml version ="1.0" encoding="UTF-8" standalone="no" ?>
<!DOCTYPE fmresultset PUBLIC "-//FMI//DTD fmresultset//EN" "http://localhost:16020/fmi/xml/fmresultset.dtd">
<fmresultset xmlns="http://www.filemaker.com/xml/fmresultset" version="1.0">
<error code="0">
</error>
<product build="11/11/2014" name="FileMaker Web Publishing Engine" version="13.0.5.518">
</product>
I use the following python to extract some of the tag names:
doc = etree.fromstring(resulttxt)
print( doc.attrib)
print(doc.tag)
print(doc[4][0][0].tag)
if(doc[4][0][0].tag == 'field'):
print 'hi'
What I'm getting though is:
{'version': '1.0'}
{http://www.filemaker.com/xml/fmresultset}fmresultset
{http://www.filemaker.com/xml/fmresultset}field
The xmlns doesn't show up as an attribute of the root tag but it is there.
And it is placed in front of each tag name which makes it difficult to loop through and use conditionals. I want doc.tag just to show the tag and not the namespace and the tag.
This is day 1 for me using this. could anyone help out?
You need to handle namespaces, in your case an empty one:
from lxml import etree as ET
data = """<?xml version ="1.0" encoding="UTF-8" standalone="no" ?>
<!DOCTYPE fmresultset PUBLIC "-//FMI//DTD fmresultset//EN" "http://localhost:16020/fmi/xml/fmresultset.dtd">
<fmresultset xmlns="http://www.filemaker.com/xml/fmresultset" version="1.0">
<error code="0">
</error>
<product build="11/11/2014" name="FileMaker Web Publishing Engine" version="13.0.5.518">
</product>
</fmresultset>
"""
namespaces = {
"myns": "http://www.filemaker.com/xml/fmresultset"
}
tree = ET.fromstring(data)
print tree.find("myns:product", namespaces=namespaces).attrib.get("name")
Prints:
FileMaker Web Publishing Engine
I'm just trying to get some values of an XML file, but I can't get anything.
<?xml version="1.0" ?>
<clasif>
<estacion>Estacion</estacion>
<ncamaras>1</ncamaras>
<fph>1</fph>
<tratamiento>3</tratamiento>
<finicio>17/07/2011</finicio>
<ffin>17/07/2011</ffin>
<hinicio>04</hinicio>
<hfin>06</hfin>
</clasif>
This is my code to access to an value:
from lxml import etree
doc = etree.parse ( 'clasif.xml' )
root = doc.getroot()
elem= doc.xpath ('//estacion')
What am I doing wrong?