更改 lxml 中的元素命名空间

Question

With lxml , I am not sure how to properly remove the namespace of an existing element and set a new one.

For instance, I'm parsing this minimal xml file:

<myroot xmlns="http://myxml.com/somevalue">
    <child1>blabla</child1>
    <child2>blablabla</child2>
</myroot>

... and I'd like it to become:

<myroot xmlns="http://myxml.com/newvalue">
    <child1>blabla/child1>
    <child2>blablabla</child2>
</myroot>

With lxml :

from lxml import etree as ET
tree = ET.parse('myfile.xml')
root= tree.getroot()

If I inspect root :

In [7]: root
Out[7]: <Element {http://myxml.com/somevalue}myroot at 0x7f6e13832588>
In [8]: root.nsmap
Out[8]: {None: 'http://myxml.com/somevalue'}
In [11]: root.tag
Out[11]: '{http://myxml.com/somevalue}myroot'

Ideally, I would like to end up with:

In [8]: root.nsmap
Out[8]: {None: 'http://myxml.com/newvalue'}
In [11]: root.tag
Out[11]: '{http://myxml.com/newvalue}myroot'

As for the tag, it's just a matter of setting the right string. How about nsmap ?

Answer 1

I agree with mzjn and Parfait; I'd use XSLT to change the namespace.

You can make the XSLT fairly reusable by having the old and new namespaces passed in as parameters.

Example...

XML Input (input.xml)

<myroot xmlns="http://myxml.com/somevalue">
    <child1>blabla</child1>
    <child2>blablabla</child2>
</myroot>

XSLT 1.0 (test.xsl)

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="yes"/>
  <xsl:strip-space elements="*"/>

  <xsl:param name="orig_namespace"/>
  <xsl:param name="new_namespace"/>

  <xsl:template match="@*|node()">
    <xsl:copy>
      <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="*" priority="1">
    <xsl:choose>
      <xsl:when test="namespace-uri()=$orig_namespace">
        <xsl:element name="{name()}" namespace="{$new_namespace}">
          <xsl:apply-templates select="@*|node()"/>
        </xsl:element>
      </xsl:when>
      <xsl:otherwise>
        <xsl:copy>
          <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
      </xsl:otherwise>
    </xsl:choose>
  </xsl:template>

</xsl:stylesheet>

Python

from lxml import etree

tree = etree.parse("input.xml")
xslt = etree.parse("test.xsl")

orig_namespace = "http://myxml.com/somevalue"
new_namespace = "http://myxml.com/newvalue"

new_tree = tree.xslt(xslt, orig_namespace=f"'{orig_namespace}'",
                     new_namespace=f"'{new_namespace}'")
print(etree.tostring(new_tree, pretty_print=True).decode("utf-8"))

Output

<myroot xmlns="http://myxml.com/newvalue">
  <child1>blabla</child1>
  <child2>blablabla</child2>
</myroot>

Also, if you use the following input (that uses a namespace prefix)...

<ns1:myroot xmlns:ns1="http://myxml.com/somevalue">
    <ns1:child1>blabla</ns1:child1>
    <ns1:child2>blablabla</ns1:child2>
</ns1:myroot>

you get this output...

<ns1:myroot xmlns:ns1="http://myxml.com/newvalue">
  <ns1:child1>blabla</ns1:child1>
  <ns1:child2>blablabla</ns1:child2>
</ns1:myroot>

See https://lxml.de/xpathxslt.html for more info on using XSLT with lxml.