I am creating a new xml file from scratch using python and the lxml library.
<route xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://www.xxxx" version="1.1"
xmlns:stm="http://xxxx/1/0/0"
xsi:schemaLocation="http://xxxx/1/0/0 stm_extensions.xsd">
I need to include this namespace information into the root tag as attributes of the route tag.
I can´t include the information into the root declaration.
from lxml import etree
root = etree.Element("route",
xmlns:xsi = "http://www.w3.org/2001/XMLSchema-instance",
xmlns = "http://www.xxxxx",
version = "1.1",
xmlns: stm = "http://xxxxx/1/0/0"
)
there is a SyntaxError: invalid syntax
How can I do that ?
Here is how it can be done:
from lxml import etree
attr_qname = etree.QName("http://www.w3.org/2001/XMLSchema-instance", "schemaLocation")
nsmap = {None: "http://www.xxxx",
"stm": "http://xxxx/1/0/0",
"xsi": "http://www.w3.org/2001/XMLSchema-instance"}
root = etree.Element("route",
{attr_qname: "http://xxxx/1/0/0 stm_extensions.xsd"},
version="1.1",
nsmap=nsmap)
print etree.tostring(root)
Output from this code (line breaks have been added for readability):
<route xmlns:stm="http://xxxx/1/0/0"
xmlns="http://www.xxxx"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xxxx/1/0/0 stm_extensions.xsd"
version="1.1"/>
The main "trick" is to use QName to create the xsi:schemaLocation attribute. An attribute with a colon in its name cannot be used as the name of a keyword argument.
I've added the declaration of the xsi prefix to nsmap, but it can actually be omitted. lxml defines default prefixes for some well-known namespace URIs, including xsi for http://www.w3.org/2001/XMLSchema-instance.
Related
In my S1000D xml, it specifies a DOCTYPE with a reference to a public URL that contains references to a number of other files that contain all the valid character entities. I've used xml.etree.ElementTree and lxml to try to parse it and get a parse error with both indicating:
undefined entity −: line 82, column 652
Even though − is a valid entity according to the ENTITY Reference specfied.
The xml top is as follow:
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE dmodule [
<!ENTITY % ISOEntities PUBLIC 'ISO 8879-1986//ENTITIES ISO Character Entities 20030531//EN//XML' 'http://www.s1000d.org/S1000D_4-1/ent/ISOEntities'>
%ISOEntities;]>
If you go out and get http://www.s1000d.org/S1000D_4-1/ent/ISOEntities, it will include 20 other ent files with one called iso-tech.ent which contains the line:
<!ENTITY minus "−"> <!-- MINUS SIGN -->
in line 82 of the xml file near column 652 is the following:
....Refer to 70−41....
How can I run a python script to parse this file without get the undefined entity?
Sorry I don't want to specify parser.entity['minus'] = chr(2212) for example. I did that for a quick fix but there are many character entity references.
I would like the parser to check Entity reference that is specified in the xml.
I'm surprised but I've gone around the sun and back and haven't found how to do this (or maybe I have but couldn't follow it).
if I update my xml file and add
<!ENTITY minus "−">
It won't fail, so It's not the xml.
It fails on the parse. Here's code I use for ElementTree
fl = os.path.join(pth, fn)
try:
root = ET.parse(fl)
except ParseError as p:
print("ParseError : ", p)
Here's the code I use for lxml
fl = os.path.join(pth, fn)
try:
parser = etree.XMLParser(load_dtd=True, resolve_entities=True)
root = etree.parse(fl, parser=parser)
except etree.XMLSyntaxError as pe:
print("lxml XMLSyntaxError: ", pe)
I would like the parser to load the ENTITY reference so that it knows that − and all the other character entities specified in all the files are valid entity characters.
Thank you so much for your advice and help.
I'm going to answer for lxml. No reason to consider ElementTree if you can use lxml.
I think the piece you're missing is no_network=False in the XMLParser; it's True by default.
Example...
XML Input (test.xml)
<!DOCTYPE doc [
<!ENTITY % ISOEntities PUBLIC 'ISO 8879-1986//ENTITIES ISO Character Entities 20030531//EN//XML' 'http://www.s1000d.org/S1000D_4-1/ent/ISOEntities'>
%ISOEntities;]>
<doc>
<test>Here's a test of minus: −</test>
</doc>
Python
from lxml import etree
parser = etree.XMLParser(load_dtd=True,
no_network=False)
tree = etree.parse("test.xml", parser=parser)
etree.dump(tree.getroot())
Output
<doc>
<test>Here's a test of minus: −</test>
</doc>
If you wanted the entity reference retained, add resolve_entities=False to the XMLParser.
Also, instead of going out to an external location to resolve the parameter entity, consider setting up an XML Catalog. This will let you resolve public and/or system identifiers to local versions.
Example using same XML input above...
XML Catalog ("catalog.xml" in the directory "catalog test" (space used in directory name for testing))
<!DOCTYPE catalog PUBLIC "-//OASIS//DTD XML Catalogs V1.1//EN" "http://www.oasis-open.org/committees/entity/release/1.1/catalog.dtd">
<catalog xmlns="urn:oasis:names:tc:entity:xmlns:xml:catalog">
<!-- The path in #uri is relative to this file (catalog.xml). -->
<uri name="http://www.s1000d.org/S1000D_4-1/ent/ISOEntities" uri="./ents/ISOEntities_stackoverflow.ent"/>
</catalog>
Entity File ("ISOEntities_stackoverflow.ent" in the directory "catalog test/ents". Changed the value to "BAM!" for testing)
<!ENTITY minus "BAM!">
Python (Changed no_network to True for additional evidence that the local version of http://www.s1000d.org/S1000D_4-1/ent/ISOEntities is being used.)
import os
from urllib.request import pathname2url
from lxml import etree
# The XML_CATALOG_FILES environment variable is used by libxml2 (which is used by lxml).
# See http://xmlsoft.org/catalog.html.
try:
xcf_env = os.environ['XML_CATALOG_FILES']
except KeyError:
# Path to catalog must be a url.
catalog_path = f"file:{pathname2url(os.path.join(os.getcwd(), 'catalog test/catalog.xml'))}"
# Temporarily set the environment variable.
os.environ['XML_CATALOG_FILES'] = catalog_path
parser = etree.XMLParser(load_dtd=True,
no_network=True)
tree = etree.parse("test.xml", parser=parser)
etree.dump(tree.getroot())
Output
<doc>
<test>Here's a test of minus: BAM!</test>
</doc>
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.
I'm trying to create xml with lxml.etree module for python2. It would be an easy task if not requirement that output should looks like:
<spring:beans xmlns="http://membrane-soa.org/proxies/1/"
xmlns:spring="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-4.2.xsd
http://membrane-soa.org/proxies/1/ http://membrane-soa.org/schemas/proxies-1.xsd">
any suggestion how can I do that? All I was able to achieve at this moment is:
<ns0:beans xmlns:ns0="http://membrane-soa.org/proxies/1/"/>
so how to have "spring" instead of "ns0"
Thanks
Use map to declare the namespaces and use None as the key specifically for default namespace :
from lxml import etree as ET
nsmap = { None: "http://membrane-soa.org/proxies/1/",
"spring": "http://www.springframework.org/schema/beans",
"xsi": "http://www.w3.org/2001/XMLSchema-instance" }
root = ET.Element("{%s}beans" % nsmap["spring"], nsmap=nsmap)
root.set("{%s}schemaLocation" % nsmap["xsi"],
"http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-4.2.xsd")
result : (after formatting)
<spring:beans
xmlns:spring="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://membrane-soa.org/proxies/1/"
spring:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-4.2.xsd"
/>
I'm using ElementTree with Python to parse an XML file to find the contents of a subchild
This is the XML file I'm trying to parse:
<?xml version='1.0' encoding='UTF-8'?>
<nvd xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://nvd.nist.gov/feeds/cve/1.2" nvd_xml_version="1.2" pub_date="2016-02-10" xsi:schemaLocation="http://nvd.nist.gov/feeds/cve/1.2 http://nvd.nist.gov/schema/nvdcve_1.2.1.xsd">
<entry type="CVE" name="CVE-1999-0001" seq="1999-0001" published="1999-12-30" modified="2010-12-16" severity="Medium" CVSS_version="2.0" CVSS_score="5.0" CVSS_base_score="5.0" CVSS_impact_subscore="2.9" CVSS_exploit_subscore="10.0" CVSS_vector="(AV:N/AC:L/Au:N/C:N/I:N/A:P)">
<desc>
<descript source="cve">ip_input.c in BSD-derived TCP/IP implementations allows remote attackers to cause a denial of service (crash or hang) via crafted packets.</descript>
</desc>
<loss_types>
<avail/>
</loss_types>
<range>
<network/>
</range>
<refs>
<ref source="OSVDB" url="http://www.osvdb.org/5707">5707</ref>
<ref source="CONFIRM" url="http://www.openbsd.org/errata23.html#tcpfix">http://www.openbsd.org/errata23.html#tcpfix</ref>
</refs>
this is my code:
import xml.etree.ElementTree as ET
if __name__ == '__main__':
tree = ET.parse('nvdcve-modified.xml')
root = tree.getroot()
print root.find('entry')
print root[0].find('desc')
the output for both lines in None
Your XML has default namespace defined at the root element level :
xmlns="http://nvd.nist.gov/feeds/cve/1.2"
Descendant elements without prefix inherits ancestor's default namespace implicitly. To find element in namespace, you can map a prefix to the namespace URI and use the prefix like so :
ns = {'d': 'http://nvd.nist.gov/feeds/cve/1.2'}
root.find('d:entry', ns)
or use the namespace URI directly :
root.find('{http://nvd.nist.gov/feeds/cve/1.2}entry')
I've noticed that python ElementTree module, changes the xml data in the following simple example :
import xml.etree.ElementTree as ET
tree = ET.parse("./input.xml")
tree.write("./output.xml")
I wouldn't expect it to change, as I've done simple read and write test without any modification. however, the results shows a different story, especially in the namespace indices (nonage --> ns0 , d3p1 --> ns1 , i --> ns2 ) :
input.xml:
<?xml version="1.0" encoding="utf-8"?>
<ServerData xmlns:i="http://www.a.org" xmlns="http://schemas.xxx/2004/07/Server.Facades.ImportExport">
<CreationDate>0001-01-01T00:00:00</CreationDate>
<Processes>
<Processes xmlns:d3p1="http://schemas.datacontract.org/2004/07/Management.Interfaces">
<d3p1:ProtectedProcess>
<d3p1:Description>/Applications/Safari.app/Contents/MacOS/Safari</d3p1:Description>
<d3p1:DiscoveredMachine i:nil="true" />
<d3p1:Id>0</d3p1:Id>
<d3p1:Name>/applications/safari.app/contents/macos/safari</d3p1:Name>
<d3p1:Path>/Applications/Safari.app/Contents/MacOS/Safari</d3p1:Path>
<d3p1:ProcessHashes xmlns:d5p1="http://schemas.datacontract.org/2004/07/Management.Interfaces.WildFire" />
<d3p1:Status>1</d3p1:Status>
<d3p1:Type>Protected</d3p1:Type>
</d3p1:ProtectedProcess>
</Processes>
</Processes>
and output.xml:
<ns0:ServerData xmlns:ns0="http://schemas.xxx/2004/07/Server.Facades.ImportExport" xmlns:ns1="http://schemas.datacontract.org/2004/07/Management.Interfaces" xmlns:ns2="http://www.a.org">
<ns0:CreationDate>0001-01-01T00:00:00</ns0:CreationDate>
<ns0:Processes>
<ns0:Processes>
<ns1:ProtectedProcess>
<ns1:Description>/Applications/Safari.app/Contents/MacOS/Safari</ns1:Description>
<ns1:DiscoveredMachine ns2:nil="true" />
<ns1:Id>0</ns1:Id>
<ns1:Name>/applications/safari.app/contents/macos/safari</ns1:Name>
<ns1:Path>/Applications/Safari.app/Contents/MacOS/Safari</ns1:Path>
<ns1:ProcessHashes />
<ns1:Status>1</ns1:Status>
<ns1:Type>Protected</ns1:Type>
</ns1:ProtectedProcess>
</ns0:Processes>
</ns0:Processes>
You would need to register the namespaces for your xml as well as their prefixes with ElementTree before reading/writing the xml using ElementTree.register_namespace function. Example -
import xml.etree.ElementTree as ET
ET.register_namespace('','http://schemas.xxx/2004/07/Server.Facades.ImportExport')
ET.register_namespace('i','http://www.a.org')
ET.register_namespace('d3p1','http://schemas.datacontract.org/2004/07/Management.Interfaces')
tree = ET.parse("./input.xml")
tree.write("./output.xml")
Without this ElementTree creates its own prefixes for the corresponding namespaces, which is what happens for your case.
This is given in the documentation -
xml.etree.ElementTree.register_namespace(prefix, uri)
Registers a namespace prefix. The registry is global, and any existing mapping for either the given prefix or the namespace URI will be removed. prefix is a namespace prefix. uri is a namespace uri. Tags and attributes in this namespace will be serialized with the given prefix, if at all possible.
(Emphasis mine)