【发布时间】:2014-08-05 07:06:16
【问题描述】:
我需要得到这个xml:
<s:Envelope xmlns:a="http://www.w3.org/2005/08/addressing" xmlns:s="http://www.w3.or/2003/05/soap-envelope">
<s:Header>
<a:Action s:mustUnderstand="1">Action</a:Action>
</s:Header>
</s:Envelope>
据我了解
from lxml.etree import Element, SubElement, QName, tostring
class XMLNamespaces:
s = 'http://www.w3.org/2003/05/soap-envelope'
a = 'http://www.w3.org/2005/08/addressing'
root = Element(QName(XMLNamespaces.s, 'Envelope'), nsmap={'s':XMLNamespaces.s, 'a':XMLNamespaces.a})
header = SubElement(root, QName(XMLNamespaces.s, 'Header'))
action = SubElement(header, QName(XMLNamespaces.a, 'Action'))
action.attrib['mustUnderstand'] = "1"
action.text = 'Action'
print tostring(root, pretty_print=True)
结果:
<s:Envelope xmlns:a="http://www.w3.org/2005/08/addressing" xmlns:s="http://www.w3.org/2003/05/soap-envelope">
<s:Header>
<a:Action mustUnderstand="1">http://schemas.xmlsoap.org/ws/2004/09/transfer/Create</a:Action>
</s:Header>
</s:Envelope>
我们可以看到,“mustUnderstand”属性前面没有命名空间前缀。那么有没有可能用lxml得到“s: mustUnderstand”呢?如果是,那么如何?
【问题讨论】:
标签: python xml namespaces lxml