lxml etree xmlparser remove unwanted namespace
I have an xml doc that I am trying to parse using Etree.lxml
<Envelope xmlns="http://www.example.com/zzz/yyy">
<Header>
<Version>1</Version>
</Header>
<Body>
some stuff
<Body>
<Envelope>
My code is:
path = "path to xml file"
from lxml import etree as ET
parser = ET.XMLParser(ns_clean=True)
dom = ET.parse(path, parser)
dom.getroot()
When I try to get dom.getroot() I get:
<Element {http://www.example.com/zzz/yyy}Envelope at 28adacac>
However I only want:
<Element Envelope at 28adacac>
When i do
dom.getroot().find("Body")
I get nothing returned. However, when I
dom.getroot().find("{http://www.example.com/zzz/yyy}Body")
I get a result.
I thought passing ns_clean=True to the parser would prevent this.
Any ideas?
import io
import lxml.etree as ET
content='''
<Envelope xmlns="http://www.example.com/zzz/yyy">
<Header>
<Version>1</Version>
</Header>
<Body>
some stuff
</Body>
</Envelope>
'''
dom = ET.parse(io.BytesIO(content))
You can find namespace-aware nodes using the xpath method:
body=dom.xpath('//ns:Body',namespaces={'ns':'http://www.example.com/zzz/yyy'})
print(body)
# [<Element {http://www.example.com/zzz/yyy}Body at 90b2d4c>]
If you really want to remove namespaces, you could use an XSL transformation:
# http://wiki.tei-c.org/index.php/Remove-Namespaces.xsl
xslt='''<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="no"/>
<xsl:template match="/|comment()|processing-instruction()">
<xsl:copy>
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>
<xsl:template match="*">
<xsl:element name="{local-name()}">
<xsl:apply-templates select="@*|node()"/>
</xsl:element>
</xsl:template>
<xsl:template match="@*">
<xsl:attribute name="{local-name()}">
<xsl:value-of select="."/>
</xsl:attribute>
</xsl:template>
</xsl:stylesheet>
'''
xslt_doc=ET.parse(io.BytesIO(xslt))
transform=ET.XSLT(xslt_doc)
dom=transform(dom)
Here we see the namespace has been removed:
print(ET.tostring(dom))
# <Envelope>
# <Header>
# <Version>1</Version>
# </Header>
# <Body>
# some stuff
# </Body>
# </Envelope>
So you can now find the Body node this way:
print(dom.find("Body"))
# <Element Body at 8506cd4>
Try using Xpath:
dom.xpath("//*[local-name() = 'Body']")
Taken (and simplified) from this page, under "The xpath() method" section
The last solution from https://bitbucket.org/olauzanne/pyquery/issue/17 can help you to avoid namespaces with little effort
apply xml.replace(' xmlns:', ' xmlnamespace:') to your xml before using pyquery so lxml will ignore namespaces
In your case, try xml.replace(' xmlns="', ' xmlnamespace="') . However, you might need something more complex if the string is expected in the bodies as well.