python - 如何从 lxml 树中剥离命名空间？

Question

继 Removing child elements in XML using python ...

感谢@Tichodroma，我有这个代码:

如果您可以使用 lxml ， 试试这个:

 import lxml.etree

 tree = lxml.etree.parse("leg.xml")
 for dog in tree.xpath("//Leg1:Dog",
                       namespaces={"Leg1": "http://what.not"}):
     parent = dog.xpath("..")[0]
     parent.remove(dog)
     parent.text = None
 tree.write("leg.out.xml")

现在 leg.out.xml看起来像这样:

 <?xml version="1.0"?>
 <Leg1:MOR xmlns:Leg1="http://what.not" oCount="7">
   <Leg1:Order>
     <Leg1:CTemp id="FO">
       <Leg1:Group bNum="001" cCount="4"/>
       <Leg1:Group bNum="002" cCount="4"/>
     </Leg1:CTemp>
     <Leg1:CTemp id="GO">
       <Leg1:Group bNum="001" cCount="4"/>
       <Leg1:Group bNum="002" cCount="4"/>
     </Leg1:CTemp>
   </Leg1:Order>
 </Leg1:MOR>

如何修改我的代码以删除 Leg1:来自所有元素标签名称的命名空间前缀？

Answer 1

从每个元素中删除命名空间前缀的一种可能方法:

def strip_ns_prefix(tree):
    #iterate through only element nodes (skip comment node, text node, etc) :
    for element in tree.xpath('descendant-or-self::*'):
        #if element has prefix...
        if element.prefix:
            #replace element name with its local name
            element.tag = etree.QName(element).localname
    return tree

另一个版本在 xpath 中进行命名空间检查而不是使用 if陈述 :

def strip_ns_prefix(tree):
    #xpath query for selecting all element nodes in namespace
    query = "descendant-or-self::*[namespace-uri()!='']"
    #for each element returned by the above xpath query...
    for element in tree.xpath(query):
        #replace element name with its local name
        element.tag = etree.QName(element).localname
    return tree