I have the following xml which i was trying to parse and wanted to get the value of attributes of root node i.e. xmlns:n1 value. But i am getting the key error using the following value.
<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<n1:Level-1C_Tile_ID xmlns:n1="https://psd-12.sentinel2.eo.esa.int/PSD/S2_PDI_Level-1C_Tile_Metadata.xsd" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="https://psd-12.sentinel2.eo.esa.int/PSD/S2_PDI_Level-1C_Tile_Metadata.xsd /dpc/app/s2ipf/FORMAT_METADATA_TILE_L1C/02.10.02/scripts/../../../schemas/02.12.05/PSD/S2_PDI_Level-1C_Tile_Metadata.xsd">
<n1:General_Info>
<TILE_ID metadataLevel="Brief">S2A_OPER_MSI_L1C_TL_MTI__20161111T091803_A007252_T43QBA_N02.04</TILE_ID>
<DATASTRIP_ID metadataLevel="Standard">S2A_OPER_MSI_L1C_DS_MTI__20161111T091803_S20161111T053350_N02.04</DATASTRIP_ID>
<DOWNLINK_PRIORITY metadataLevel="Standard">NOMINAL</DOWNLINK_PRIORITY>
<SENSING_TIME metadataLevel="Standard">2016-11-11T05:33:50.068Z</SENSING_TIME>
<Archiving_Info metadataLevel="Expertise">
<ARCHIVING_CENTRE>MTI_</ARCHIVING_CENTRE>
<ARCHIVING_TIME>2016-11-11T10:53:22.600451Z</ARCHIVING_TIME>
</Archiving_Info>
</n1:General_Info>
</n1:Level-1C_Tile_ID>
Code :
from lxml import etree
tree = etree.parse(XML_FILE_CONTENT_PASTED_ABOVE)
root = tree.getroot()
print(root.attrib['xmlns:n1'])
Error :
KeyError: 'xmlns:n1'
Desired output :
https://psd-12.sentinel2.eo.esa.int/PSD/S2_PDI_Level-1C_Tile_Metadata.xsd
A namespace declaration (xmlns:n1='...') looks like an attribute, but it is not part of the attrib dictionary of an element.
To get the namespace URI associated with the n1 prefix, use nsmap:
print(root.nsmap["n1"])
Related
I am trying to parse an XML file in Python with the built in xml module and Elemnt tree, but what ever I try to do according to the documentation, it does not give me what I need.
I am trying to extract all the value tags into a list
<?xml version="1.0" encoding="UTF-8"?>
<CustomField xmlns="http://soap.sforce.com/2006/04/metadata">
<fullName>testPicklist__c</fullName>
<externalId>false</externalId>
<label>testPicklist</label>
<required>false</required>
<trackFeedHistory>false</trackFeedHistory>
<type>Picklist</type>
<valueSet>
<restricted>true</restricted>
<valueSetDefinition>
<sorted>false</sorted>
<value>
<fullName>a 32</fullName>
<default>false</default>
<label>a 32</label>
</value>
<value>
<fullName>23 432;:</fullName>
<default>false</default>
<label>23 432;:</label>
</value>
and here is the example code that I cant get to work. It's very basic and all I have issues is the xpath.
from xml.etree.ElementTree import ElementTree
field_filepath= "./testPicklist__c.field-meta.xml"
mydoc = ElementTree()
mydoc.parse(field_filepath)
root = mydoc.getroot()
print(root.findall(".//value")
print(root.findall(".//*/value")
print(root.findall("./*/value")
Since the root element has attribute xmlns="http://soap.sforce.com/2006/04/metadata", every element in the document will belong to this namespace. So you're actually looking for {http://soap.sforce.com/2006/04/metadata}value elements.
To search all <value> elements in this document you have to specify the namespace argument in the findall() function
from xml.etree.ElementTree import ElementTree
field_filepath= "./testPicklist__c.field-meta.xml"
mydoc = ElementTree()
mydoc.parse(field_filepath)
root = mydoc.getroot()
# get the namespace of root
ns = root.tag.split('}')[0][1:]
# create a dictionary with the namespace
ns_d = {'my_ns': ns}
# get all the values
values = root.findall('.//my_ns:value', namespaces=ns_d)
# print the values
for value in values:
print(value)
Outputs:
<Element '{http://soap.sforce.com/2006/04/metadata}value' at 0x7fceea043ba0>
<Element '{http://soap.sforce.com/2006/04/metadata}value' at 0x7fceea043e20>
Alternatively you can just search for the {http://soap.sforce.com/2006/04/metadata}value
# get all the values
values = root.findall('.//{http://soap.sforce.com/2006/04/metadata}value')
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 have an xml like shown below
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE dtbook PUBLIC "-//INFO//INFO info 2005-3//EN" "http://url">
<dtbook xmlns="http://www.daisy.org/z3986/2005/dtbook/" version="2005-3" xml:lang="ml">
<head>....
</dtbook>
I open the file like so,
with open("filename.xml") as f:
tree = ET.parse(f)
root = tree.getroot()
When I try to get the root tag, I get,
print(root.tag)
{http://www.daisy.org/z3986/2005/dtbook/}dtbook
whereas if I remove all the attributes from the root tag i.e. dtbook, I get the correct output i.e. dtbook
print(root.tag)
dtbook
I cannot remove the attributes. Is there a way to get this working without removing the attributes??
This is called a namespace and is supposed to be in front. You can simply remove the namespace by splitting your string at {}
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 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')