I need to get some info after a specific tag in lxml.
the xml doc looks like this
<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/
ns/j2ee/web-app_2_4.xsd"
version="2.4">
<display-name>Community Bank</display-name>
<description>WebGoat for Cigital</description>
<context-param>
<param-name>PropertiesPath</param-name>
<param-value>/WEB-INF/properties.txt</param-value>
<description>This is the path to the properties file from the servlet root</description>
</context-param>
<servlet>
<servlet-name>Index</servlet-name>
<servlet-class>com.cigital.boi.servlet.index</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Index</servlet-name>
<url-pattern>/index</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>Index</servlet-name>
<url-pattern>/index.html</url-pattern>
</servlet-mapping>
I want to read com.cigital.boi.servlet.index .
I have used this code to read everything under servlets
context = etree.parse(handle)
list = parser.xpath('//servlet')
print list
list contains nothing
more info : iterating over the context field i found these lines.
<Element {http://java.sun.com/xml/ns/j2ee}servlet-name at 2ad19e6eca48>
<Element {http://java.sun.com/xml/ns/j2ee}servlet-class at 2ad19e6ecaf8>
I am thinking as I have not included name space while searching , output is empty list.
please suggest hoe to read "com.cigital.boi.servlet.index" in the servlet-class tag
Try following:
from lxml import etree
context = etree.parse(handle)
print next(x.text for x in context.xpath('.//*[local-name()="servlet-class"]'))
Alternative:
from lxml import etree
context = etree.parse(handle)
nsmap = context.getroot().nsmap.copy()
nsmap['xmlns'] = nsmap.pop(None)
print next(x.text for x in context.xpath('.//xmlns:servlet-class', namespaces=nsmap))
Related
The output of my xml parsing is not es expected.
The xml file
<?xml version="1.0"?>
<stationaer xsi:schemaLocation="http:/foo.bar" xmlns="http://foo.bar" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<einrichtung>
<name>Name</name>
</einrichtung>
<einrichtung>
<name>Name</name>
</einrichtung>
</stationaer>
I would expect to get something like root.tag == 'stationaer' and child.tag = 'einrichtung'.
See the outpout at the end.
This is the MWE
#!/usr/bin/env python3
import pathlib
import lxml
from lxml import etree
import pandas
xml_src = '''<?xml version="1.0"?>
<stationaer xsi:schemaLocation="http:/foo.bar" xmlns="http://foo.bar" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<einrichtung>
<name>Name</name>
</einrichtung>
<einrichtung>
<name>Name</name>
</einrichtung>
</stationaer>
'''
# tree = etree.parse(file_path)
# root = tree.getroot()
root = etree.fromstring(xml_src)
print(repr(root.tag))
print(repr(root.text))
child = root.getchildren()[0]
print(repr(child.tag))
print(repr(child.text))
The output for root is
'{http://foo.bar}stationaer'
'\n '
and for child
'{http://foo.bar}einrichtung'
'\n '
I don't understand what's going on here and why that URL is in the output.
This is actually not unexpected. The elements in the XML document are bound to the http://foo.bar default namespace. The namespace is declared by xmlns="http://foo.bar" on the root element and the declaration is inherited by all descendants.
The special notation with the namespace URI enclosed in curly braces ({http://foo.bar}stationaer) is never used in XML documents, but it is used by lxml and ElementTree when printing element (tag) names. It can also be used when searching or creating elements that belong to a namespace.
More information:
https://www.w3.org/TR/xml-names/
https://lxml.de/tutorial.html#namespaces
https://docs.python.org/3/library/xml.etree.elementtree.html#parsing-xml-with-namespaces
I am newbie on Python programming. I have requirement where I need to read the xml structure and build the new soap request xml by adding namespace like here is the example what I have
Below XML which i get from other system:
<foo>
<bar>
<type foobar="1"/>
<type foobar="2"/>
</bar>
</foo>
I want final result like below
<?xml version="1.0"?>
<soa:foo xmlns:soa="https://www.w3schools.com/furniture">
<soa:bar>
<soa:type foobar="1"/>
<soa:type foobar="2"/>
</soa:bar>
</soa:foo>
I tried to look in python document but not able to find
One option is to use lxml to iterate over all of the elements and add the namespace uri to the .tag property.
You can use register_namespace() to bind the uri to the desired prefix.
Example...
from lxml import etree
tree = etree.parse("input.xml")
etree.register_namespace("soa", "https://www.w3schools.com/furniture")
for elem in tree.iter():
elem.tag = f"{{https://www.w3schools.com/furniture}}{elem.tag}"
print(etree.tostring(tree, pretty_print=True).decode())
Printed output...
<soa:foo xmlns:soa="https://www.w3schools.com/furniture">
<soa:bar>
<soa:type foobar="1"/>
<soa:type foobar="2"/>
</soa:bar>
</soa:foo>
I am trying to parse an XML document using lxml etree. The XML doc I am parsing looks like this:
<?xml version="1.0" encoding="UTF-8"?>
<metadata xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://www.openarchives.org/OAI/2.0/">\t
<codeBook version="2.5" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="ddi:codebook:2_5" xsi:schemaLocation="ddi:codebook:2_5 http://www.ddialliance.org/Specification/DDI-Codebook/2.5/XMLSchema/codebook.xsd">
<docDscr>
<citation>
<titlStmt>
<titl>Test Title</titl>
</titlStmt>
<prodStmt>
<prodDate/>
</prodStmt>
</citation>
</docDscr>
<stdyDscr>
<citation>
<titlStmt>
<titl>Test Title 2</titl>
<IDNo agency="UKDA">101</IDNo>
</titlStmt>
<rspStmt>
<AuthEnty>TestAuthEntry</AuthEnty>
</rspStmt>
<prodStmt>
<copyright>Yes</copyright>
</prodStmt>
<distStmt/>
<verStmt>
<version date="">1</version>
</verStmt>
</citation>
<stdyInfo>
<subject>
<keyword>2009</keyword>
<keyword>2010</keyword>
<topcClas>CLASS</topcClas>
<topcClas>ffdsf</topcClas>
</subject>
<abstract>This is an abstract piece of text.</abstract>
<sumDscr>
<timePrd event="single">2020</timePrd>
<nation>UK</nation>
<anlyUnit>Test</anlyUnit>
<universe>test</universe>
<universe>hello</universe>
<dataKind>fdsfdsf</dataKind>
</sumDscr>
</stdyInfo>
<method>
<dataColl>
<timeMeth>test timemeth</timeMeth>
<dataCollector>test data collector</dataCollector>
<sampProc>test sampprocess</sampProc>
<deviat>test deviat</deviat>
<collMode>test collMode</collMode>
<sources/>
</dataColl>
</method>
<dataAccs>
<setAvail>
<accsPlac>Test accsPlac</accsPlac>
</setAvail>
<useStmt>
<restrctn>NONE</restrctn>
</useStmt>
</dataAccs>
<othrStdyMat>
<relPubl>122</relPubl>
<relPubl>12332</relPubl>
</othrStdyMat>
</stdyDscr>
</codeBook>
</metadata>
I wrote the following code to try and process it:
from lxml import etree
import pdb
f = open('/vagrant/out2.xml', 'r')
xml_str = f.read()
xml_doc = etree.fromstring(xml_str)
f.close()
From what I understand from the lxml xpath docs, I should be able to get the text from a specific element as follows:
xml_doc.xpath('/metadata/codeBook/docDscr/citation/titlStmt/titl/text()')
However, when I run this it returns an empty array.
The only xpath I can get to return something is using a wildcard:
xml_doc.xpath('*')
Which returns [<Element {ddi:codebook:2_5}codeBook at 0x7f8da8a413f8>].
I've read through the docs and I'm not understanding what is going wrong with this. Any help is appreciated.
You need to take the default namespace into account so instead of
xml_doc.xpath('/metadata/codeBook/docDscr/citation/titlStmt/titl/text()')
use
xml_doc.xpath.xpath(
'/oai:metadata/ddi:codeBook/ddi:docDscr/ddi:citation/ddi:titlStmt/ddi:titl/text()',
namespaces={
'oai': 'http://www.openarchives.org/OAI/2.0/',
'ddi': 'ddi:codebook:2_5'
}
)
I am parsing an XML output from VCloud, however I am not able to reach to the values
<?xml version="1.0" encoding="UTF-8"?>
<SupportedVersions xmlns="http://www.vmware.com/vcloud/versions" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.vmware.com/vcloud/versions http://10.10.6.12/api/versions/schema/versions.xsd">
<VersionInfo>
<Version>1.5</Version>
<LoginUrl>https://api.vcd.portal.skyscapecloud.com/api/sessions</LoginUrl>
<MediaTypeMapping>
<MediaType>application/vnd.vmware.vcloud.instantiateVAppTemplateParams+xml</MediaType>
<ComplexTypeName>InstantiateVAppTemplateParamsType</ComplexTypeName>
<SchemaLocation>http://api.vcd.portal.skyscapecloud.com/api/v1.5/schema/master.xsd</SchemaLocation>
</MediaTypeMapping>
<MediaTypeMapping>
<MediaType>application/vnd.vmware.admin.vmwProviderVdcReferences+xml</MediaType>
<ComplexTypeName>VMWProviderVdcReferencesType</ComplexTypeName>
<SchemaLocation>http://api.vcd.portal.skyscapecloud.com/api/v1.5/schema/vmwextensions.xsd</SchemaLocation>
</MediaTypeMapping>
<MediaTypeMapping>
<MediaType>application/vnd.vmware.vcloud.customizationSection+xml</MediaType>
<ComplexTypeName>CustomizationSectionType</ComplexTypeName>
<SchemaLocation>http://api.vcd.portal.skyscapecloud.com/api/v1.5/schema/master.xsd</SchemaLocation>
</MediaTypeMapping>
this is what I have been using
import xml.etree.ElementTree as ET
data = ET.fromstring(content)
versioninfo = data.findall("VersionInfo/Version")
print len(versioninfo)
print versioninfo.text
however this gives a blank output...any suggestions?
Try this:
import xml.etree.ElementTree as ET
data = ET.fromstring(content)
versioninfo = data.find(
"ns:VersionInfo/ns:Version",
namespaces={'ns':'http://www.vmware.com/vcloud/versions'})
print versioninfo.text
Use .find(), not .findall() to return a single element
Your XML uses namespaces. The full path to your desired object is: '{http://www.vmware.com/vcloud/versions}VersionInfo/{http://www.vmware.com/vcloud/versions}Version' By passing in the namespaces parameter, you are able to use the shortcut syntax: ns:VersionInfo/ns:Version.
I try to parse a huge file. The sample is below. I try to take <Name>, but I can't
It works only without this string
<LevelLayout xmlns="http://schemas.datacontract.org/2004/07/ArcherTech.Common.Domain" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
xml2 = '''<?xml version="1.0" encoding="UTF-8"?>
<PackageLevelLayout>
<LevelLayouts>
<LevelLayout levelGuid="4a54f032-325e-4988-8621-2cb7b49d8432">
<LevelLayout xmlns="http://schemas.datacontract.org/2004/07/ArcherTech.Common.Domain" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
<LevelLayoutSectionBase>
<LevelLayoutItemBase>
<Name>Tracking ID</Name>
</LevelLayoutItemBase>
</LevelLayoutSectionBase>
</LevelLayout>
</LevelLayout>
</LevelLayouts>
</PackageLevelLayout>'''
from lxml import etree
tree = etree.XML(xml2)
nodes = tree.xpath('/PackageLevelLayout/LevelLayouts/LevelLayout[#levelGuid="4a54f032-325e-4988-8621-2cb7b49d8432"]/LevelLayout/LevelLayoutSectionBase/LevelLayoutItemBase/Name')
print nodes
Your nested LevelLayout XML document uses a namespace. I'd use:
tree.xpath('.//LevelLayout[#levelGuid="4a54f032-325e-4988-8621-2cb7b49d8432"]//*[local-name()="Name"]')
to match the Name element with a shorter XPath expression (ignoring the namespace altogether).
The alternative is to use a prefix-to-namespace mapping and use those on your tags:
nsmap = {'acd': 'http://schemas.datacontract.org/2004/07/ArcherTech.Common.Domain'}
tree.xpath('/PackageLevelLayout/LevelLayouts/LevelLayout[#levelGuid="4a54f032-325e-4988-8621-2cb7b49d8432"]/acd:LevelLayout/acd:LevelLayoutSectionBase/acd:LevelLayoutItemBase/acd:Name',
namespaces=nsmap)
lxml's xpath method has a namespaces parameter. You can pass it a dict mapping namespace prefixes to namespaces. Then you can refer build XPaths that use the namespace prefix:
xml2 = '''<?xml version="1.0" encoding="UTF-8"?>
<PackageLevelLayout>
<LevelLayouts>
<LevelLayout levelGuid="4a54f032-325e-4988-8621-2cb7b49d8432">
<LevelLayout xmlns="http://schemas.datacontract.org/2004/07/ArcherTech.Common.Domain" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
<LevelLayoutSectionBase>
<LevelLayoutItemBase>
<Name>Tracking ID</Name>
</LevelLayoutItemBase>
</LevelLayoutSectionBase>
</LevelLayout>
</LevelLayout>
</LevelLayouts>
</PackageLevelLayout>'''
namespaces={'ns': 'http://schemas.datacontract.org/2004/07/ArcherTech.Common.Domain',
'i': 'http://www.w3.org/2001/XMLSchema-instance'}
import lxml.etree as ET
# This is an lxml.etree._Element, not a tree, so don't call it tree
root = ET.XML(xml2)
nodes = root.xpath(
'''/PackageLevelLayout/LevelLayouts/LevelLayout[#levelGuid="4a54f032-325e-4988-8621-2cb7b49d8432"]
/ns:LevelLayout/ns:LevelLayoutSectionBase/ns:LevelLayoutItemBase/ns:Name''', namespaces = namespaces)
print nodes
yields
[<Element {http://schemas.datacontract.org/2004/07/ArcherTech.Common.Domain}Name at 0xb74974dc>]