lxml etree xmlparser remove unwanted namespace

2019-01-06 10:56发布

站内问答 / Python
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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?

4条回答
你好瞎i
2楼-- · 2019-01-06 11:18

Try using Xpath:

dom.xpath("//*[local-name() = 'Body']")

Taken (and simplified) from this page, under "The xpath() method" section

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Luminary・发光体
3楼-- · 2019-01-06 11:28

You're showing the result of the repr() call. When you programmatically move through the tree, you can simply choose to ignore the namespace.

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何必那么认真
4楼-- · 2019-01-06 11:31 
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>
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Fickle 薄情
5楼-- · 2019-01-06 11:37

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.

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